Merge pull request #4 from AndreasFMueller/master

update

184 files changed, 19768 insertions, 3606 deletions

diff --git a/buch/chapters/010-potenzen/tschebyscheff.tex b/buch/chapters/010-potenzen/tschebyscheff.tex
index ccc2e97..6d21a68 100644
--- a/buch/chapters/010-potenzen/tschebyscheff.tex
+++ b/buch/chapters/010-potenzen/tschebyscheff.tex
@@ -102,7 +102,7 @@ die Sütztstellen so zu wählen, dass $l(x)$ kleine Funktionswerte hat.
 Stützstellen in gleichen Abständen erweisen sich dafür als ungeeignet,
 da $l(x)$ nahe $x_0$ und $x_n$ sehr stark oszilliert.
 
-\subsection{Definition der Tschebyscheff-Polynome}
+\subsection{Definition der Tschebyscheff-Polynome \label{sub:definiton_der_tschebyscheff-Polynome}}
 \begin{figure}
 \centering
 \includegraphics[width=\textwidth]{chapters/010-potenzen/images/lissajous.pdf}
@@ -199,6 +199,7 @@ T_0(x)=1.
 \end{equation}
 Damit können die Tschebyscheff-Polynome sehr effizient berechnet werden:
 \begin{equation}
+\label{eq:tschebyscheff-polynome}
 \begin{aligned}
 T_0(x)
 &=1
diff --git a/buch/chapters/030-geometrie/hyperbolisch.tex b/buch/chapters/030-geometrie/hyperbolisch.tex
index 2938316..d2d0da2 100644
--- a/buch/chapters/030-geometrie/hyperbolisch.tex
+++ b/buch/chapters/030-geometrie/hyperbolisch.tex
@@ -163,9 +163,9 @@ In der speziellen Relativitätstheorie spielt das Minkowski-Skalarprodukt
 eine besondere Rolle.
 Die Koordinaten $x_0$ hat darin die Bedeutung der Zeit,
 man weiss aus Experimenten wie dem Michelson-Morley-Experiment,
-dass die Grösse $\langle x,x\rangle$ ist eine Invariante ist.
+dass die Grösse $\langle x,x\rangle$ eine Invariante ist.
 Die Transformationen mit der Matrix $A$ beschreiben also zulässige
-Koordinatentransformationenn, die Invariante erhalten.
+Koordinatentransformationen, die Invariante erhalten.
 
 Für Transformationen, die zusätzlich die Zeitrichtung erhalten sollen,
 muss $a_{00}=a_{11}=c>0$ verlangt werden.
@@ -174,7 +174,7 @@ muss $a_{00}=a_{11}=c>0$ verlangt werden.
 Unter der Annahme $c>0$ lässt sich die Matrix vollständig
 durch den Parameter $t=s/c$ beschreiben.
 Dividiert man \eqref{buch:geometrie:hyperbolish:eqn:cs} durch $c^2$,
-kann $c$ durch $t$ ausdrücken:
+kann man $c$ durch $t$ ausdrücken:
 \[
 \frac{1}{c^2}
 =
@@ -199,10 +199,10 @@ H_t
 t&1
 \end{pmatrix}.
 \]
-Diese Formeln erinnern natürlich and die Formeln, mit denen
+Diese Formeln erinnern natürlich an die Formeln, mit denen
 der hyperbolische Sinus und Kosinus aus dem hyperbolischen
-Tangens berechnet werden kann.
-Dieser Zusammenhang und soll im nächsten Abschnitt hergestellt
+Tangens berechnet werden können.
+Dieser Zusammenhang soll im nächsten Abschnitt hergestellt
 werden.
 
 %
diff --git a/buch/chapters/070-orthogonalitaet/gaussquadratur.tex b/buch/chapters/070-orthogonalitaet/gaussquadratur.tex
index a5af7d2..c7dfb31 100644
--- a/buch/chapters/070-orthogonalitaet/gaussquadratur.tex
+++ b/buch/chapters/070-orthogonalitaet/gaussquadratur.tex
@@ -20,7 +20,7 @@ Ein solches Polynom $p(x)$ hat $n+1$ Koeffizienten, die aus dem
 linearen Gleichungssystem der $n+1$ Gleichungen $p(x_i)=f(x_i)$ 
 ermittelt werden können.
 
-Das Interpolationspolynom $p(x)$ lässt sich abera uch direkt 
+Das Interpolationspolynom $p(x)$ lässt sich aber auch direkt 
 angeben.
 Dazu konstruiert man zuerst die Polynome
 \[
diff --git a/buch/chapters/070-orthogonalitaet/orthogonal.tex b/buch/chapters/070-orthogonalitaet/orthogonal.tex
index df04514..793b78d 100644
--- a/buch/chapters/070-orthogonalitaet/orthogonal.tex
+++ b/buch/chapters/070-orthogonalitaet/orthogonal.tex
@@ -641,6 +641,7 @@ H_w
 f\colon(a,b) \to \mathbb{R}
 \;\bigg|\;
 \int_a^b |f(x)|^2 w(x)\,dx
+<\infty
 \biggr\}.
 \]
 Die Funktionen $f\in H_w$ haben folgende Eigenschaften
diff --git a/buch/chapters/070-orthogonalitaet/sturm.tex b/buch/chapters/070-orthogonalitaet/sturm.tex
index 742ec0a..80bd5f4 100644
--- a/buch/chapters/070-orthogonalitaet/sturm.tex
+++ b/buch/chapters/070-orthogonalitaet/sturm.tex
@@ -15,7 +15,7 @@ Skalarproduktes selbstadjungierten Operators erkannt wurden.
 %
 % Differentialgleichungen
 %
-\subsection{Differentialgleichung}
+\subsection{Differentialgleichung \label{sub:differentailgleichung}}
 Das klassische Sturm-Liouville-Problem ist das folgende Eigenwertproblem.
 Gesucht sind Lösungen der Differentialgleichung
 \begin{equation}
@@ -405,7 +405,7 @@ L
 %
 % Beispiele
 %
-\subsection{Beispiele}
+\subsection{Beispiele\label{sub:beispiele_sturm_liouville_problem}}
 Die meisten der früher vorgestellten Funktionenfamilien stellen sich
 als Lösungen eines geeigneten Sturm-Liouville-Problems heraus.
 Alle Eigenschaften aus der Sturm-Liouville-Theorie gelten daher
diff --git a/buch/chapters/075-fourier/bessel.tex b/buch/chapters/075-fourier/bessel.tex
index 7e978f7..db7880b 100644
--- a/buch/chapters/075-fourier/bessel.tex
+++ b/buch/chapters/075-fourier/bessel.tex
@@ -454,7 +454,8 @@ Terme mit $\pm n$ können wegen
 \[
 \left.
 \begin{aligned}
-J_{-n}(\xi) &= (-1)^n J_n(\xi) 
+J_{-n}(\xi) &= (-1)^n J_n(\xi)
+\label{buch:fourier:eqn:symetrie}
 \\
 i^{-n}&=(-1)^n i^n
 \end{aligned}
diff --git a/buch/chapters/110-elliptisch/uebungsaufgaben/1.tex b/buch/chapters/110-elliptisch/uebungsaufgaben/1.tex
index af094c6..2d08e56 100644
--- a/buch/chapters/110-elliptisch/uebungsaufgaben/1.tex
+++ b/buch/chapters/110-elliptisch/uebungsaufgaben/1.tex
@@ -25,7 +25,7 @@ Auslenkung.
 Formulieren Sie den Energieerhaltungssatz für die Gesamtenergie $E$
 dieses Oszillators.
 Leiten Sie daraus eine nichtlineare Differentialgleichung erster Ordnung
-for den anharmonischen Oszillator ab, die sie in der Form
+für den anharmonischen Oszillator ab, die sie in der Form
 $\frac12m\dot{x}^2 = f(x)$ schreiben.
 \item
 Die Amplitude der Schwingung ist derjenige $x$-Wert, für den die
diff --git a/buch/papers/0f1/images/konvergenzNegativ.pdf b/buch/papers/0f1/images/konvergenzNegativ.pdf
index 03b2ba1..07d2a44 100644
--- a/buch/papers/0f1/images/konvergenzNegativ.pdf
+++ b/buch/papers/0f1/images/konvergenzNegativ.pdf
diff --git a/buch/papers/0f1/images/konvergenzPositiv.pdf b/buch/papers/0f1/images/konvergenzPositiv.pdf
index 2e45129..8e1e7e4 100644
--- a/buch/papers/0f1/images/konvergenzPositiv.pdf
+++ b/buch/papers/0f1/images/konvergenzPositiv.pdf
diff --git a/buch/papers/0f1/listings/kettenbruchIterativ.c b/buch/papers/0f1/listings/kettenbruchIterativ.c
index d897b8f..3caaf43 100644
--- a/buch/papers/0f1/listings/kettenbruchIterativ.c
+++ b/buch/papers/0f1/listings/kettenbruchIterativ.c
@@ -1,53 +1,27 @@
 /**

-  * @brief Calculates the Hypergeometric Function 0F1(;b;z)

-  * @param b0 in 0F1(;b0;z)

-  * @param z in 0F1(;b0;z)

-  * @param n number of itertions (precision)

+  * @brief Calculates the Hypergeometric Function 0F1(;c;z)

+  * @param c in 0F1(;c;z)

+  * @param z in 0F1(;c;z)

+  * @param k number of itertions (precision)

   * @return Result

   */

-static double fractionRekursion0f1(const double c, const double z, unsigned int n)

+static double fractionIter0f1(const double c, const double z, unsigned int k)

 {

     //declaration

     double a = 0.0;

     double b = 0.0;

-    double Ak = 0.0;

-    double Bk = 0.0;

-    double Ak_1 = 0.0;

-    double Bk_1 = 0.0;

-    double Ak_2 = 0.0;

-    double Bk_2 = 0.0;

+    double abk = 0.0;

+    double temp = 0.0;

 

-    for (unsigned int k = 0; k <= n; ++k)

+    for (; k > 0; --k)

     {

-        if (k == 0)

-        {

-            a = 1.0; //a0

-            //recursion fomula for A0, B0

-            Ak = a;

-            Bk = 1.0;

-        }

-        else if (k == 1)

-        {

-            a = 1.0; //a1

-            b = z/c; //b1

-            //recursion fomula for A1, B1

-            Ak = a * Ak_1 + b * 1.0;

-            Bk = a * Bk_1;

-        }

-        else

-        {

-            a = 1 + (z / (k * ((k - 1) + c)));//ak

-            b = -(z / (k * ((k - 1) + c)));   //bk

-            //recursion fomula for Ak, Bk

-            Ak = a * Ak_1 + b * Ak_2;

-            Bk = a * Bk_1 + b * Bk_2;

-        }        

-        //save old values

-        Ak_2 = Ak_1;

-        Bk_2 = Bk_1;

-        Ak_1 = Ak;

-        Bk_1 = Bk; 

+        abk = z / (k * ((k - 1) + c));  //abk = ak, bk

+        

+        a = k > 1 ? (1 + abk) : 1;      //a0, a1

+        b = k > 1 ? -abk : abk;         //b1

+

+        temp = b / (a + temp);          //bk / (ak + last result)

     }

-    //approximation fraction

-    return  Ak/Bk;

-}

+

+    return a + temp;                    //a0 + temp

+}
\ No newline at end of file
diff --git a/buch/papers/0f1/listings/kettenbruchRekursion.c b/buch/papers/0f1/listings/kettenbruchRekursion.c
index 3caaf43..d897b8f 100644
--- a/buch/papers/0f1/listings/kettenbruchRekursion.c
+++ b/buch/papers/0f1/listings/kettenbruchRekursion.c
@@ -1,27 +1,53 @@
 /**

-  * @brief Calculates the Hypergeometric Function 0F1(;c;z)

-  * @param c in 0F1(;c;z)

-  * @param z in 0F1(;c;z)

-  * @param k number of itertions (precision)

+  * @brief Calculates the Hypergeometric Function 0F1(;b;z)

+  * @param b0 in 0F1(;b0;z)

+  * @param z in 0F1(;b0;z)

+  * @param n number of itertions (precision)

   * @return Result

   */

-static double fractionIter0f1(const double c, const double z, unsigned int k)

+static double fractionRekursion0f1(const double c, const double z, unsigned int n)

 {

     //declaration

     double a = 0.0;

     double b = 0.0;

-    double abk = 0.0;

-    double temp = 0.0;

+    double Ak = 0.0;

+    double Bk = 0.0;

+    double Ak_1 = 0.0;

+    double Bk_1 = 0.0;

+    double Ak_2 = 0.0;

+    double Bk_2 = 0.0;

 

-    for (; k > 0; --k)

+    for (unsigned int k = 0; k <= n; ++k)

     {

-        abk = z / (k * ((k - 1) + c));  //abk = ak, bk

-        

-        a = k > 1 ? (1 + abk) : 1;      //a0, a1

-        b = k > 1 ? -abk : abk;         //b1

-

-        temp = b / (a + temp);          //bk / (ak + last result)

+        if (k == 0)

+        {

+            a = 1.0; //a0

+            //recursion fomula for A0, B0

+            Ak = a;

+            Bk = 1.0;

+        }

+        else if (k == 1)

+        {

+            a = 1.0; //a1

+            b = z/c; //b1

+            //recursion fomula for A1, B1

+            Ak = a * Ak_1 + b * 1.0;

+            Bk = a * Bk_1;

+        }

+        else

+        {

+            a = 1 + (z / (k * ((k - 1) + c)));//ak

+            b = -(z / (k * ((k - 1) + c)));   //bk

+            //recursion fomula for Ak, Bk

+            Ak = a * Ak_1 + b * Ak_2;

+            Bk = a * Bk_1 + b * Bk_2;

+        }        

+        //save old values

+        Ak_2 = Ak_1;

+        Bk_2 = Bk_1;

+        Ak_1 = Ak;

+        Bk_1 = Bk; 

     }

-

-    return a + temp;                    //a0 + temp

-}
\ No newline at end of file
+    //approximation fraction

+    return  Ak/Bk;

+}

diff --git a/buch/papers/0f1/teil0.tex b/buch/papers/0f1/teil0.tex
index adccac7..335cf92 100644
--- a/buch/papers/0f1/teil0.tex
+++ b/buch/papers/0f1/teil0.tex
@@ -5,11 +5,11 @@
 %

 \section{Ausgangslage\label{0f1:section:ausgangslage}}

 \rhead{Ausgangslage}

-Die Hypergeometrische Funktion $\mathstrut_0F_1$ wird in vielen Funktionen als Basisfunktion benutzt, 

-zum Beispiel um die Airy Funktion zu berechnen. 

+Die hypergeometrische Funktion $\mathstrut_0F_1$ wird in vielen Funktionen als Basisfunktion benutzt, 

+zum Beispiel um die Airy-Funktion zu berechnen. 

 In der GNU Scientific Library \cite{0f1:library-gsl} 

 ist die Funktion $\mathstrut_0F_1$ vorhanden. 

-Allerdings wirft die Funktion, bei negativen Übergabenwerten wie zum Beispiel \verb+gsl_sf_hyperg_0F1(1, -1)+, eine Exception. 

+Allerdings wirft die Funktion bei negativen Übergabewerten wie zum Beispiel \verb+gsl_sf_hyperg_0F1(1, -1)+ eine Exception. 

 Bei genauerer Untersuchung hat sich gezeigt, dass die Funktion je nach Betriebssystem funktioniert oder eben nicht. 

 So kann die Funktion unter Windows fehlerfrei aufgerufen werden, beim Mac OS und Linux sind negative Übergabeparameter im Moment nicht möglich.

-Ziel dieser Arbeit war es zu evaluieren, ob es mit einfachen mathematischen Operationen möglich ist, die Hypergeometrische Funktion $\mathstrut_0F_1$ zu implementieren.

+Ziel dieser Arbeit war es zu evaluieren, ob es mit einfachen mathematischen Operationen möglich ist, die hypergeometrische Funktion $\mathstrut_0F_1$ zu implementieren.

diff --git a/buch/papers/0f1/teil1.tex b/buch/papers/0f1/teil1.tex
index 2ca9647..8d00f95 100644
--- a/buch/papers/0f1/teil1.tex
+++ b/buch/papers/0f1/teil1.tex
@@ -6,12 +6,11 @@
 \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

-beschrieben.

+Basierend auf den Herleitungen des 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 ein Speziallfall der allgemein definierten Funktion $\mathstrut_pF_q$.

 

 \begin{definition}

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

@@ -42,7 +41,8 @@ Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$
     \mathstrut_0F_1

     \biggl(

     \begin{matrix}

-    \\

+    \text{---}

+    \\\

     b_1

     \end{matrix}

     ;

@@ -58,24 +58,24 @@ Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$
 

 

 

-\subsection{Airy Funktion

+\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 $\operatorname{Ai}(x)$ und die verwandte Funktion $\operatorname{Bi}(x)$ 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, ergeben sich wie in Abschnitt \ref{buch:differentialgleichungen:section:hypergeometrisch} hergeleitet, folgende Lösungen der Airy-Differentialgleichung zu den Anfangsbedingungen $\operatorname{Ai}(0)=1$ und $\operatorname{Ai}'(0)=0$, sowie $\operatorname{Bi}(0)=0$ und $\operatorname{Bi}'(0)=1$:

 

 \begin{align}

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

-Ai(x)

-=

+\operatorname{Ai}(x)

+=&

 \sum_{k=0}^\infty

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

 =

@@ -83,8 +83,8 @@ Ai(x)
 \begin{matrix}\text{---}\\\frac23\end{matrix};\frac{x^3}{9}

 \biggr).

 \\

-Bi(x)

-=

+\operatorname{Bi}(x)

+=&

 \sum_{k=0}^\infty

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

 =

@@ -95,7 +95,6 @@ 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 dieser Arbeit die Airy Funktion $\operatorname{Ai}(x)$ benutzt.

 

 

diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex
index 9269961..fdcb0fc 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, die in vollständiger Form auf Github \cite{0f1:code} zu finden sind. 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

 

 \begin{align}

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

@@ -23,41 +23,84 @@ Die naheliegendste Lösung ist die Programmierung der Potenzreihe. Allerdings is
     \frac{1}{c}

     +\frac{z^1}{(c+1) \cdot 1}

     + \cdots

-    + \frac{z^{20}}{c(c+1)(c+2)\cdots(c+19) \cdot 2.4 \cdot 10^{18}}

+    + \frac{z^{20}}{c(c+1)(c+2)\cdots(c+19) \cdot 2.4 \cdot 10^{18}}.

 \end{align}

 

 \lstinputlisting[style=C,float,caption={Potenzreihe.},label={0f1:listing:potenzreihe}, firstline=59]{papers/0f1/listings/potenzreihe.c}

 

 \subsection{Kettenbruch

 \label{0f1:subsection:kettenbruch}}

-Ein endlicher Kettenbruch ist ein Bruch der Form

+Eine weitere Variante zur Berechnung von $\mathstrut_0F_1(;c;z)$ ist die Umsetzung als Kettenbruch.

+Der Vorteil einer Umsetzung als Kettenbruch gegenüber der Potenzreihe ist die schnellere Konvergenz.

+

+\subsubsection{Grundlage}

+Ein endlicher Kettenbruch \cite{0f1:wiki-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 sind.

+

+\subsubsection{Rekursionsbeziehungen und Kettenbrüche}

+Wenn es eine Relation analytischer Funktion $f_i(z)$ hat, dann gibt es einen Kettenbruch für das Verhältnis $\frac{f_i(z)}{f_{i-1}(z)}$ \cite{0f1:wiki-fraction}. 

+Nimmt man die Gleichung

+\begin{equation*}

+	f_{i-1} - f_i = k_i z f_{i+1},

+\end{equation*}

+wo $f_i$ analytische Funktionen sind und $i > 0$ ist, sowie $k_i$ konstant.

+Ergibt sich der Zusammenhang

 \begin{equation*}

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

+	\cfrac{f_i}{f_{i-1}} = \cfrac{1}{1+k_iz\cfrac{f_{i+1}}{f_i}}.

 \end{equation*}

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

-Die Kurzschreibweise für einen allgemeinen Kettenbruch ist 

+Geht man einen Schritt weiter und nimmt für $g_i = \frac{f_i}{f_{i-1}}$ an, kommt man zur Formel

 \begin{equation*}

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

+	g_i = \cfrac{1}{1+k_izg_{i+1}}.

 \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}:

+Setzt man dies nun für $g_1$ in den Bruch ein, ergibt sich

 \begin{equation*}

+	g_1 = \cfrac{f_1}{f_0} = \cfrac{1}{1+k_izg_2} = \cfrac{1}{1+\cfrac{k_1z}{1+k_2zg_3}} = \cdots

+\end{equation*}

+Repetiert man dies unendlich, erhält man einen Kettenbruch in der Form:

+\begin{equation}

+	\label{0f1:math:rekursion:eq}

+	\cfrac{f_1}{f_0} = \cfrac{1}{1+\cfrac{k_1z}{1+\cfrac{k_2z}{1+\cfrac{k_3z}{\cdots}}}}.

+\end{equation}

+

+\subsubsection{Rekursion für $\mathstrut_0F_1$}

+Angewendet auf die Potenzreihe

+\begin{equation}

+	\label{0f1:math:potenzreihe:0f1:eq}

 	\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}

+kann durch Substitution bewiesen werden, dass

+\begin{equation*}

+	\mathstrut_0F_1(;c-1;z) - \mathstrut_0F_1(;c;z) = \frac{z}{c(c-1)} \cdot \mathstrut_0F_1(;c+1;z)

+\end{equation*}

+eine Relation dazu ist.

+Wenn man für $f_i$ und $k_i$ die Annahme

+\begin{align*}

+	f_i =& \mathstrut_0F_1(;c+i;z)\\

+	k_i	=& \frac{1}{(c+i)(c+i-1)}

+\end{align*}

+trifft und in die Formel \eqref{0f1:math:rekursion:eq} einsetzt, erhält man:

+\begin{equation*}

+	\cfrac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)} = \cfrac{1}{1+\cfrac{\cfrac{z}{c(c+1)}}{1+\cfrac{\cfrac{z}{(c+1)(c+2)}}{1+\cfrac{\cfrac{z}{(c+2)(c+3)}}{\cdots}}}}.

 \end{equation*}

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

+

+\subsubsection{Algorithmus}

+Da mit obigen Formeln nur ein Verhältnis zwischen $	\frac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)}$ berechnet wurde, braucht es weitere Relationen um $\mathstrut_0F_1(;c;z)$ zu erhalten.

+So ergeben ähnliche Relationen nach Wolfram Alpha \cite{0f1:wolfram-0f1} den 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. 

-\cite{0f1:wolfram-0f1}

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

+

 

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

 

 \subsection{Rekursionsformel

 \label{0f1:subsection:rekursionsformel}}

-Wesentlich stabiler zur Berechnung eines Kettenbruches ist die Rekursionsformel. Nachfolgend wird die verkürzte Herleitung vom Kettenbruch zur Rekursionsformel aufgezeigt. Eine vollständige Schritt für Schritt Herleitung ist im Seminarbuch Numerik, im Kapitel Kettenbrüche zu finden. \cite{0f1:kettenbrueche}

+Wesentlich stabiler zur Berechnung eines Kettenbruches ist die Rekursionsformel. Nachfolgend wird die verkürzte Herleitung vom Kettenbruch zur Rekursionsformel aufgezeigt. Eine vollständige Schritt für Schritt Herleitung ist im Seminarbuch Numerik, im Kapitel Kettenbrüche \cite{0f1:kettenbrueche} zu finden.

 

 \subsubsection{Herleitung}

 Ein Näherungsbruch in der Form

@@ -69,7 +112,7 @@ lässt sich zu
 	\cfrac{A_k}{B_k} = \cfrac{b_{k+1}}{a_{k+1} + \cfrac{p}{q}} = \frac{b_{k+1} \cdot q}{a_{k+1} \cdot q + p}

 \end{align*}

 umformen.

-Dies lässt sich auch durch die folgende Matrizenschreibweise ausdrücken:

+Dies lässt sich auch durch die Matrizenschreibweise

 \begin{equation*}

 	\begin{pmatrix}

 		A_k\\

@@ -89,11 +132,12 @@ Dies lässt sich auch durch die folgende Matrizenschreibweise ausdrücken:
 	\end{pmatrix}.

 	%\label{0f1:math:rekursionsformel:herleitung}

 \end{equation*}

+ausdrücken.

 Wendet man dies nun auf den Kettenbruch in der Form

 \begin{equation*}

 	\frac{A_k}{B_k} = a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{\cdots}{\cdots+\cfrac{b_{k-1}}{a_{k-1} + \cfrac{b_k}{a_k}}}}}

 \end{equation*}

-an, ergibt sich folgende Matrixdarstellungen:

+an, ergibt sich die Matrixdarstellungen:

 

 \begin{align*}

 	\begin{pmatrix}

@@ -117,7 +161,7 @@ an, ergibt sich folgende Matrixdarstellungen:
 	\begin{pmatrix}

 		b_k\\

 		a_k

-	\end{pmatrix}

+	\end{pmatrix}.

 \end{align*}

 Nach vollständiger Induktion ergibt sich für den Schritt $k$, die Matrix

 \begin{equation}

@@ -136,15 +180,15 @@ Nach vollständiger Induktion ergibt sich für den Schritt $k$, die Matrix
 		a_k

 	\end{pmatrix}.

 \end{equation}

-Und Schlussendlich kann der Näherungsbruch

+Und schlussendlich kann der Näherungsbruch

 \[

-\frac{Ak}{Bk}

+\frac{A_k}{B_k}

 \] 

 berechnet werden.

 

 

-\subsubsection{Lösung}

-Die Berechnung von $A_k, B_k$ \eqref{0f1:math:matrix:ende:eq} kann man auch ohne die Matrizenschreibweise aufschreiben: \cite{0f1:wiki-fraction}

+\subsubsection{Algorithmus}

+Die Berechnung von $A_k, B_k$ gemäss \eqref{0f1:math:matrix:ende:eq} kann man auch ohne die Matrizenschreibweise \cite{0f1:kettenbrueche} aufschreiben:

 \begin{itemize}

 \item Startbedingungen:

 \begin{align*}

@@ -163,10 +207,10 @@ B_{k+1} &= B_{k-1} \cdot b_k + B_k \cdot a_k
 \end{aligned}

 \]

 \item

-Näherungsbruch: \qquad$\displaystyle\frac{A_k}{B_k}$

+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 \eqref{0f1:listing:kettenbruchRekursion} ist, dass im Vergleich zum Code (Listing \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..147668a 100644
--- a/buch/papers/0f1/teil3.tex
+++ b/buch/papers/0f1/teil3.tex
@@ -6,59 +6,54 @@
 \section{Auswertung

 \label{0f1:section:teil3}}

 \rhead{Resultate}

-Im Verlauf des Seminares hat sich gezeigt, 

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

+Im Verlauf dieser Arbeit hat sich gezeigt, 

+das einen einfachen mathematischen 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 \cite{0f1:wolfram-0f1} ab.

 

 \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.

+Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass nach drei Iterationen ($k = 3$) die Funktionen genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel 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 gerechnet werden, ist wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. Bei der Rekursionsformel muss beachtet werden, dass sie zwar erst nach 35 Approximationen gänzlich konvergiert, allerdings nach 27 Iterationen sich nicht mehr gross verändert.

+

+Ist $z$ negativ, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies aufgrund des Vorzeichens zu alternierenden Termen. So steigt bei allen Algorithmen zuerst die Differenz zum erwarteten Endwert. Erst nach genügend Iterationen sind die Terme so klein, dass sie das Endresultat nicht mehr signifikant beeinflussen. Während die Potenzreihe zusammen mit dem Kettenbruch nach 34 Approximationen konvergiert, braucht die Rekursionsformel noch zwei Iterationen mehr.  

 

-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.

 

 \subsection{Stabilität

 \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.

-

-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.

+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 (Listing \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 (Listing \ref{0f1:listing:potenzreihe}) das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät im Nenner. Dies führt zu einer Bereichsüberschreitung des \verb+double+ Bereiches \cite{0f1:double}, der spätesten ab $k=167$ eintritt. 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 Konvergenz zu gewährleisten, muss wie in Abbildung \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 Potenz von $z$, was zum Phänomen der Auslöschung \cite{0f1:SeminarNumerik} führt. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind nach Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} die Potenzreihe, der Kettenbruch, als auch die Rekursionsformel, bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da alle Algorithmen auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Diese programmiertechnischen Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} und \ref{0f1:ausblick:plot:konvergenz:negativ} festzustellen.

 

 \begin{figure}

     \centering

     \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzAiry.pdf}

-    \caption{Konvergenz nach drei Iterationen, dargestellt anhand der Airy Funktion zu den Anfangsbedingungen $Ai(0)=1$ und $Ai'(0)=0$.

+    \caption{Konvergenz nach drei Iterationen, dargestellt anhand der Airy Funktion zu den Anfangsbedingungen $\operatorname{Ai}(0)=1$ und $\operatorname{Ai}'(0)=0$.

     \label{0f1:ausblick:plot:airy:konvergenz}}

 \end{figure}

 

 \begin{figure}

     \centering

     \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzPositiv.pdf}

-    \caption{Konvergenz: Logarithmisch dargestellte Differenz vom erwarteten Endresultat.

+    \caption{Konvergenz mit positivem $z$; Logarithmisch dargestellter absoluter Fehler.

     \label{0f1:ausblick:plot:konvergenz:positiv}}

 \end{figure}

 

 \begin{figure}

     \centering

     \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzNegativ.pdf}

-    \caption{Konvergenz: Logarithmisch dargestellte Differenz vom erwarteten Endresultat.

+    \caption{Konvergenz mit negativem $z$; Logarithmisch dargestellter absoluter Fehler.

     \label{0f1:ausblick:plot:konvergenz:negativ}}

 \end{figure}

 

 \begin{figure}

     \centering

     \includegraphics[width=1\textwidth]{papers/0f1/images/stabilitaet.pdf}

-    \caption{Stabilität der 3 Algorithmen verglichen mit der Referenz Funktion $Ai(x)$.

+    \caption{Stabilität der drei Algorithmen verglichen mit der Referenz Funktion $\operatorname{Ai}(x)$.

     \label{0f1:ausblick:plot:airy:stabilitaet}}

 \end{figure}

 

diff --git a/buch/papers/ellfilter/einleitung.tex b/buch/papers/ellfilter/einleitung.tex
index 37fd89f..ae7127f 100644
--- a/buch/papers/ellfilter/einleitung.tex
+++ b/buch/papers/ellfilter/einleitung.tex
@@ -1,56 +1,73 @@
 \section{Einleitung}
 
-% Lineare filter
-
-% Filter, Signalverarbeitung
-
-
-Der womöglich wichtigste Filtertyp ist das Tiefpassfilter.
-Dieses soll im Durchlassbereich unter der Grenzfrequenz $\Omega_p$ unverstärkt durchlassen und alle anderen Frequenzen vollständig auslöschen.
-
-% Bei der Implementierung von Filtern
-
-In der Elektrotechnik führen Schaltungen mit linearen Bauelementen wie Kondensatoren, Spulen und Widerständen immer zu linearen zeitinvarianten Systemen (LTI-System von englich \textit{time-invariant system}).
-Die Übertragungsfunktion im Frequenzbereich $|H(\Omega)|$ eines solchen Systems ist dabei immer eine rationale Funktion, also eine Division von zwei Polynomen.
-Die Polynome habe dabei immer reelle oder komplex-konjugierte Nullstellen.
-
-
-\begin{equation} \label{ellfilter:eq:h_omega}
-    | H(\Omega)|^2 = \frac{1}{1 + \varepsilon_p^2 F_N^2(w)}, \quad w=\frac{\Omega}{\Omega_p}
+Filter sind womöglich eines der wichtigsten Elementen in der Signalverarbeitung und finden Anwendungen in der digitalen und analogen Elektrotechnik.
+Besonders hilfreich ist die Untergruppe der linearen Filter.
+Elektronische Schaltungen mit linearen Bauelementen wie Kondensatoren, Spulen und Widerständen führen immer zu linearen zeitinvarianten Systemen (LTI-System von englich \textit{time-invariant system}).
+Durch die Linearität werden beim das Filtern keine neuen Frequenzanteile erzeugt, was es erlaubt, einen Frequenzanteil eines Signals verzerrungsfrei herauszufiltern. %TODO review sentence
+Diese Eigenschaft macht es Sinnvoll, lineare Filter im Frequenzbereich zu beschreiben.
+Die Übertragungsfunktion eines linearen Filters im Frequenzbereich $H(\Omega)$ ist dabei immer eine rationale Funktion, also ein Quotient von zwei Polynomen.
+Dabei ist $\Omega = 2 \pi f$ die Frequenzeinheit.
+Die Polynome haben dabei immer reelle oder komplex-konjugierte Nullstellen.
+
+Ein breit angewendeter Filtertyp ist das Tiefpassfilter, welches beabsichtigt alle Frequenzen eines Signals oberhalb der Grenzfrequenz $\Omega_p$ auszulöschen.
+Der Rest soll dabei unverändert passieren.
+Aus dem Tiefpassifilter können dann durch Transformationen auch Hochpassfilter, Bandpassfilter und Bandsperren realisiert werden.
+Ein solches Filter hat idealerweise die Frequenzantwort
+\begin{equation}
+    H(\Omega) =
+    \begin{cases}
+        1  & \Omega < \Omega_p \\
+        0  & \Omega < \Omega_p
+    \end{cases},
 \end{equation}
-
-$\Omega = 2 \pi f$ ist die analoge Frequenz
-
-
-% Linear filter
-Damit das Filter implementierbar und stabil ist, muss $H(\Omega)^2$ eine rationale Funktion sein, deren Nullstellen und Pole auf der linken Halbebene liegen.
-
-$N \in \mathbb{N} $ gibt dabei die Ordnung des Filters vor, also die maximale Anzahl Pole oder Nullstellen.
-
-Damit ein Filter die Passband Kondition erfüllt muss $|F_N(w)| \leq 1 \forall |w| \leq 1$ und für $|w| \geq 1$ sollte die Funktion möglichst schnell divergieren.
-Eine einfaches Polynom, dass das erfüllt, erhalten wir wenn $F_N(w) = w^N$.
+wie dargestellt in Abbildung \ref{ellfilter:fig:lp}
+\begin{figure}
+    \centering
+    \input{papers/ellfilter/tikz/filter.tikz.tex}
+    \caption{Frequenzantwort eines Tiefpassfilters.}
+    \label{ellfilter:fig:lp}
+\end{figure}
+Leider ist eine solche Funktion nicht als rationale Funktion darstellbar.
+Aus diesem Grund sind realisierbare Approximationen gesucht.
+Jede Approximation wird einen kontinuierlichen Übergang zwischen Durchlassbereich und Sperrbereich aufweisen.
+Oft wird dabei der Faktor $1/\sqrt{2}$ als Schwelle zwischen den beiden Bereichen gewählt.
+Somit lassen sich lineare Tiefpassfilter mit folgender Funktion zusammenfassen:
+\begin{equation}
+    | H(\Omega)|^2 = \frac{1}{1 + \varepsilon_p^2 F_N^2(w)}, \quad w=\frac{\Omega}{\Omega_p},
+\end{equation}
+wobei $F_N(w)$ eine rationale Funktion ist, $|F_N(w)| \leq 1 ~\forall~ |w| \leq 1$ erfüllt und für $|w| \geq 1$ möglichst schnell divergiert.
+Des weiteren müssen alle Nullstellen und Pole von $F_N$ auf der linken Halbebene liegen, damit das Filter implementierbar und stabil ist.
+$w$ ist die normalisierte Frequenz, die es erlaubt ein Filter unabhängig von der Grenzfrequenz zu beschrieben.
+Bei $w=1$ hat das Filter eine Dämpfung von $1/(1+\varepsilon^2)$.
+$N \in \mathbb{N} $ gibt die Ordnung des Filters vor, also die maximale Anzahl Pole oder Nullstellen.
+Je hoher $N$ gewählt wird, desto steiler ist der Übergang in denn Sperrbereich.
+Grössere $N$ sind erfordern jedoch aufwendigere Implementierungen und haben mehr Phasenverschiebung.
+Eine einfache Funktion, die für $F_N$ eingesetzt werden kann, ist das Polynom $w^N$.
 Tatsächlich erhalten wir damit das Butterworth Filter, wie in Abbildung \ref{ellfilter:fig:butterworth} ersichtlich.
 \begin{figure}
     \centering
     \input{papers/ellfilter/python/F_N_butterworth.pgf}
-    \caption{$F_N$ für Butterworth filter. Der grüne Bereich definiert die erlaubten Werte für alle $F_N$-Funktionen.}
+    \caption{$F_N$ für Butterworth filter. Der grüne und gelbe Bereich definiert die erlaubten Werte für alle $F_N$-Funktionen.}
     \label{ellfilter:fig:butterworth}
 \end{figure}
-
-wenn $F_N(w)$ eine rationale Funktion ist, ist auch $H(\Omega)$ eine rationale Funktion und daher ein lineares Filter. %proof?
-
+Eine Reihe von rationalen Funktionen können für $F_N$ eingesetzt werden, um Tiefpassfilter\-approximationen mit unterschiedlichen Eigenschaften zu erhalten:
 \begin{align}
     F_N(w) & =
     \begin{cases}
         w^N                            & \text{Butterworth} \\
         T_N(w)                         & \text{Tschebyscheff, Typ 1}  \\
         [k_1 T_N (k^{-1} w^{-1})]^{-1} & \text{Tschebyscheff, Typ 2}  \\
-        R_N(w, \xi)                    & \text{Elliptisch (Cauer)}    \\
+        R_N(w, \xi)                    & \text{Elliptisch}    \\
     \end{cases}
 \end{align}
-
-Mit der Ausnahme vom Butterworth filter sind alle Filter nach speziellen Funktionen benannt.
-Alle diese Filter sind optimal für unterschiedliche Anwendungsgebiete.
+Mit der Ausnahme vom Butterworth-Filter sind alle Filter nach speziellen Funktionen benannt.
+Alle diese Filter sind optimal hinsichtlich einer Eigenschaft.
 Das Butterworth-Filter, zum Beispiel, ist maximal flach im Durchlassbereich.
-Das Tschebyscheff-1 Filter sind maximal steil für eine definierte Welligkeit im Durchlassbereich, währendem es im Sperrbereich monoton abfallend ist.
+Das Tschebyscheff-1 Filter ist maximal steil für eine definierte Welligkeit im Durchlassbereich, währendem es im Sperrbereich monoton abfallend ist.
 Es scheint so als sind gewisse Eigenschaften dieser speziellen Funktionen verantwortlich für die Optimalität dieser Filter.
+
+Dieses Paper betrachtet die Theorie hinter dem elliptischen Filter, dem wohl exotischsten dieser Auswahl.
+Es weist sich aus durch den steilsten Übergangsbereich für eine gegebene Filterdesignspezifikation.
+Des weiteren kann es als Verallgemeinerung des Tschebyscheff-Filters angesehen werden.
+
+% wenn $F_N(w)$ eine rationale Funktion ist, ist auch $H(\Omega)$ eine rationale Funktion und daher ein lineares Filter. %proof?
diff --git a/buch/papers/ellfilter/elliptic.tex b/buch/papers/ellfilter/elliptic.tex
index 88bfbfe..67bcca0 100644
--- a/buch/papers/ellfilter/elliptic.tex
+++ b/buch/papers/ellfilter/elliptic.tex
@@ -1,92 +1,100 @@
 \section{Elliptische rationale Funktionen}
 
-Kommen wir nun zum eigentlichen Teil dieses Papers, den elliptischen rationalen Funktionen
+Kommen wir nun zum eigentlichen Teil dieses Papers, den elliptischen rationalen Funktionen \cite{ellfilter:bib:orfanidis}
 \begin{align}
-    R_N(\xi, w) &= \cd \left(N~f_1(\xi)~\cd^{-1}(w, 1/\xi), f_2(\xi)\right) \\
-                &= \cd \left(N~\frac{K_1}{K}~\cd^{-1}(w, k), k_1)\right) , \quad k= 1/\xi, k_1 = 1/f(\xi) \\
+    R_N(\xi, w) &= \cd \left(N~f_1(\xi)~\cd^{-1}(w, 1/\xi), f_2(\xi)\right) \label{ellfilter:eq:elliptic}\\
+                &= \cd \left(N~\frac{K_1}{K}~\cd^{-1}(w, k), k_1\right) , \quad k= 1/\xi, k_1 = 1/f(\xi) \\
                 &= \cd \left(N~K_1~z , k_1 \right), \quad w= \cd(z K, k)
 \end{align}
-
-
-sieht ähnlich aus wie die trigonometrische Darstellung der Tschebyschef-Polynome \eqref{ellfilter:eq:chebychef_polynomials}
+Beim Betrachten dieser Definition, fällt die Ähnlichkeit zur trigonometrische Darstellung der Tsche\-byschef-Polynome \eqref{ellfilter:eq:chebychef_polynomials} auf.
 Anstelle vom Kosinus kommt hier die $\cd$-Funktion zum Einsatz.
 Die Ordnungszahl $N$ kommt auch als Faktor for.
-Zusätzlich werden noch zwei verschiedene elliptische Module $k$ und $k_1$ gebraucht.
-
-
-
-Sinus entspricht $\sn$
-
-Damit die Nullstellen an ähnlichen Positionen zu liegen kommen wie bei den Tschebyscheff-Polynomen, muss die $\cd$-Funktion gewählt werden.
+Zusätzlich werden noch zwei verschiedene elliptische Moduli $k$ und $k_1$ gebraucht.
+Bei $k = k_1 = 0$ wird der $\cd$ zum Kosinus und wir erhalten in diesem Spezialfall die Tschebyschef-Polynome.
 
+Durch das Konzept vom fundamentalen Rechteck, siehe Abbildung \ref{buch:elliptisch:fig:ellall} können für alle inversen Jacobi elliptischen Funktionen die Positionen der Null- und Polstellen anhand eines Diagramms ermittelt werden.
 Die $\cd^{-1}(w, k)$-Funktion ist um $K$ verschoben zur $\sn^{-1}(w, k)$-Funktion, wie ersichtlich in Abbildung \ref{ellfilter:fig:cd}.
 \begin{figure}
     \centering
     \input{papers/ellfilter/tikz/cd.tikz.tex}
     \caption{
-        $z$-Ebene der Funktion $z = \sn^{-1}(w, k)$.
+        $z$-Ebene der Funktion $z = \cd^{-1}(w, k)$.
         Die Funktion ist in der realen Achse $4K$-periodisch und in der imaginären Achse $2jK^\prime$-periodisch.
     }
     \label{ellfilter:fig:cd}
 \end{figure}
-Auffallend ist, dass sich alle Nullstellen und Polstellen um $K$ verschoben haben.
-
-Durch das Konzept vom fundamentalen Rechteck, siehe Abbildung \ref{ellfilter:fig:fundamental_rectangle} können für alle inversen Jaccobi elliptischen Funktionen die Positionen der Null- und Polstellen anhand eines Diagramms ermittelt werden.
-Der erste Buchstabe bestimmt die Position der Nullstelle und der zweite Buchstabe die Polstelle.
+Auffallend an der $w = \cd(z, k)$-Funktion ist, dass sich $w$ auf der reellen Achse wie der Kosinus immer zwischen $-1$ und $1$ bewegt, während bei $\mathrm{Im(z) = K^\prime}$ die Werte zwischen $\pm 1/k$ und $\pm \infty$ verlaufen.
+Die Idee des elliptischen Filter ist es, diese zwei Equirippel-Zonen abzufahren, wie ersichtlich in Abbildung \ref{ellfilter:fig:cd2}, welche Analog zu Abbildung \ref{ellfilter:fig:arccos2} gesehen werden kann.
 \begin{figure}
     \centering
-    \input{papers/ellfilter/tikz/fundamental_rectangle.tikz.tex}
+    \input{papers/ellfilter/tikz/cd2.tikz.tex}
     \caption{
-        Fundamentales Rechteck der inversen Jaccobi elliptischen Funktionen.
+        $z_1=N\frac{K_1}{K}\cd^{-1}(w, k)$-Ebene der elliptischen rationalen Funktionen.
+        Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen werden passiert.
+        Als Vereinfachung ist die Funktion nur für $w>0$ dargestellt.
     }
-    \label{ellfilter:fig:fundamental_rectangle}
+    \label{ellfilter:fig:cd2}
 \end{figure}
-
-Auffallend an der $w = \sn(z, k)$-Funktion ist, dass sich $w$ auf der reellen Achse wie der Kosinus immer zwischen $-1$ und $1$ bewegt, während bei $\mathrm{Im(z) = K^\prime}$ die Werte zwischen $\pm 1/k$ und $\pm \infty$ verlaufen.
-Die Funktion hat also Equirippel-Verhalten um $w=0$ und um $w=\pm \infty$.
-Falls es möglich ist diese Werte abzufahren im Sti der Tschebyscheff-Polynome, kann ein Filter gebaut werden, dass Equirippel-Verhalten im Durchlass- und Sperrbereich aufweist.
-
-
-
-Analog zu Abbildung \ref{ellfilter:fig:arccos2} können wir auch bei den elliptisch rationalen Funktionen die komplexe $z$-Ebene betrachten, wie ersichtlich in Abbildung \ref{ellfilter:fig:cd2}, um die besser zu verstehen.
+Das elliptische Filter hat im Gegensatz zum Tschebyscheff-Filter drei Zonen.
+Im Durchlassbereich werden wie beim Tschebyscheff-Filter die Nullstellen durchlaufen.
+Statt dass $z_1$ für alle $w>1$ in die imaginäre Richtung geht, bewegen wir uns im Sperrbereich wieder in reeller Richtung, wo Pole durchlaufen werden.
+Aus dieser Sicht kann der Sperrbereich vom Tschebyscheff-Filter als unendlich langer Übergangsbereich angesehen werden.
+% Falls es möglich ist diese Werte abzufahren im Stil der Tschebyscheff-Polynome, kann ein Filter gebaut werden, dass Equirippel-Verhalten im Durchlass- und Sperrbereich aufweist.
+Da sich die Funktion im Übergangsbereich nur zur nächsten Reihe bewegt, ist der Übergangsbereich monoton steigend.
+Theoretisch könnte eine gleiches Durchlass- und Sperrbereichverhalten erreicht werden, wenn die Funktion auf eine andere Reihe ansteigen würde.
+Dies würde jedoch zu Oszillationen zwischen $1$ und $1/k$ im Übergangsbereich führen.
+Abbildung \ref{ellfilter:fig:elliptic_freq} zeigt eine elliptisch rationale Funktion und die Frequenzantwort des daraus resultierenden Filters.
 \begin{figure}
     \centering
-    \input{papers/ellfilter/tikz/cd2.tikz.tex}
-    \caption{
-        $z_1$-Ebene der elliptischen rationalen Funktionen.
-        Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen passiert.
-    }
-    \label{ellfilter:fig:cd2}
+    \input{papers/ellfilter/python/elliptic.pgf}
+    \caption{$F_N$ und die resultierende Frequenzantwort eines elliptischen Filters.}
+    \label{ellfilter:fig:elliptic_freq}
 \end{figure}
-% Da die $\cd^{-1}$-Funktion 
-
 
+\subsection{Gradgleichung}
 
+Damit die Pol- und Nullstellen genau in dieser Konstellation durchfahren werden, müssen die elliptischen Moduli des inneren und äusseren $\cd$ aufeinander abgestimmt werden.
+In der reellen Richtung müssen sich die Periodizitäten $K$ und $K_1$ um den Faktor $N$ unterscheiden, während die imagiäre Periodizitäten $K^\prime$ und $K^\prime_1$ gleich bleiben müssen.
+Zur Erinnerung, $K$ und $K^\prime$ sind durch elliptische Integrale definiert und vom Modul $k$ abhängig wie ersichtlich in Abbildung \ref{ellfilter:fig:kprime}.
 \begin{figure}
     \centering
-    \input{papers/ellfilter/python/F_N_elliptic.pgf}
-    \caption{$F_N$ für ein elliptischs filter.}
-    \label{ellfilter:fig:elliptic}
+    \input{papers/ellfilter/python/k.pgf}
+    \caption{Die Periodizitäten in realer und imaginärer Richtung in Abhängigkeit vom elliptischen Modul $k$.}
+    \label{ellfilter:fig:kprime}
 \end{figure}
-
-\subsection{Degree Equation}
-
-Der $\cd^{-1}$ Term muss so verzogen werden, dass die umgebene $\cd$-Funktion die Nullstellen und Pole trifft.
-Dies trifft ein wenn die Degree Equation erfüllt ist.
-
+$K$ und $K^\prime$ sind durch die Ortskurve $K + jK^\prime$ aneinander gebunden und benötigen den Zusatzfaktor $K_1/K$ in \eqref{ellfilter:eq:elliptic}, um die genanten Forderungen einzuhalten.
+Abbildung \ref{ellfilter:fig:degree_eq} zeigt das Problem geometrisch auf, wobei zwei Punkte $K+jK^\prime$ und $K_1+jK_1^\prime$ auf der Ortskurve gesucht sind.
+\begin{figure}
+    \centering
+    \input{papers/ellfilter/tikz/elliptic_transform2.tikz}
+    \caption{Die Gradgleichung als geometrisches Problem ($N=3$).}
+    \label{ellfilter:fig:degree_eq}
+\end{figure}
+Algebraisch kann so die Gradgleichung
 \begin{equation}
     N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1}
 \end{equation}
+aufgestellt werden, dessen Lösung ist gegeben durch
+\begin{equation} %TODO check
+k_1 = k^N \prod_{i=1}^L \sn^4 \Bigg( \frac{2i - 1}{N} K, k \Bigg),
+\quad \text{wobei} \quad
+N = 2L+r.
+\end{equation}
+Die Herleitung ist sehr umfassend und wird in \cite{ellfilter:bib:orfanidis} im Detail angeschaut.
 
+% \begin{figure}
+%     \centering
+%     \input{papers/ellfilter/tikz/elliptic_transform1.tikz}
+%     \caption{Die Gradgleichung als geometrisches Problem.}
+% \end{figure}
 
-Leider ist das lösen dieser Gleichung nicht trivial.
-Die Rechnung wird in \ref{ellfilter:bib:orfanidis} im Detail angeschaut.
-
-
-\subsection{Polynome?}
+\subsection{Schlussfolgerung}
 
+Die elliptischen Filter können als direkte Erweiterung der Tschebyscheff-Filter verstanden werden.
 Bei den Tschebyscheff-Polynomen haben wir gesehen, dass die Trigonometrische Formel zu einfachen Polynomen umgewandelt werden kann.
-Im gegensatz zum $\cos^{-1}$ hat der $\cd^{-1}$ nicht nur Nullstellen sondern auch Pole.
+Im elliptischen Fall entstehen so rationale Funktionen mit Nullstellen und auch Pole.
 Somit entstehen bei den elliptischen rationalen Funktionen, wie es der name auch deutet, rationale Funktionen, also ein Bruch von zwei Polynomen.
 
-Da Transformationen einer rationalen Funktionen mit Grundrechenarten, wie es in \eqref{ellfilter:eq:h_omega} der Fall ist, immer noch rationale Funktionen ergeben, stellt dies kein Problem für die Implementierung dar.
+% Da Transformationen einer rationalen Funktionen mit Grundrechenarten, wie es in \eqref{ellfilter:eq:h_omega} der Fall ist, immer noch rationale Funktionen ergeben, stellt dies kein Problem für die Implementierung dar.
+
+
diff --git a/buch/papers/ellfilter/jacobi.tex b/buch/papers/ellfilter/jacobi.tex
index 6a208fa..567bbcc 100644
--- a/buch/papers/ellfilter/jacobi.tex
+++ b/buch/papers/ellfilter/jacobi.tex
@@ -1,17 +1,17 @@
 \section{Jacobische elliptische Funktionen}
 
-%TODO $z$ or $u$ for parameter?
-
-Für das elliptische Filter wird statt der, für das Tschebyscheff-Filter benutzen Kreis-Trigonometrie die elliptischen Funktionen gebraucht.
+Für das elliptische Filter werden, wie es der Name bereits deutet, elliptische Funktionen gebraucht.
+Wie die trigonometrischen Funktionen Zusammenhänge eines Kreises darlegen, beschreiben die elliptischen Funktionen Ellipsen.
+Es ist daher naheliegend, dass Kosinus des Tschebyscheff-Filters mit einem elliptischen Pendant ausgetauscht werden könnte.
 Der Begriff elliptische Funktion wird für sehr viele Funktionen gebraucht, daher ist es hier wichtig zu erwähnen, dass es ausschliesslich um die Jacobischen elliptischen Funktionen geht.
 
+Die Jacobi elliptischen Funktionen werden ausführlich im Kapitel \ref{buch:elliptisch:section:jacobi} behandelt.
 Im Wesentlichen erweitern die Jacobi elliptischen Funktionen die trigonometrische Funktionen für Ellipsen.
 Zum Beispiel gibt es analog zum Sinus den elliptischen $\sn(z, k)$.
 Im Gegensatz zum den trigonometrischen Funktionen haben die elliptischen Funktionen zwei parameter.
-Zum einen gibt es den \textit{elliptische Modul} $k$, der die Exzentrizität der Ellipse parametrisiert.
-Zum andern das Winkelargument $z$.
+Den \textit{elliptische Modul} $k$, der die Exzentrizität der Ellipse parametrisiert und das Winkelargument $z$.
 Im Kreis ist der Radius für alle Winkel konstant, bei Ellipsen ändert sich das.
-Dies hat zur Folge, dass bei einer Ellipse die Kreisbodenstrecke nicht linear zum Winkel verläuft.
+Dies hat zur Folge, dass bei einer Ellipse die Kreisbogenlänge nicht linear zum Winkel verläuft.
 Darum kann hier nicht der gewohnte Winkel verwendet werden.
 Das Winkelargument $z$ kann durch das elliptische Integral erster Art
 \begin{equation}
@@ -27,17 +27,8 @@ Das Winkelargument $z$ kann durch das elliptische Integral erster Art
             1-k^2 \sin^2 \theta
         }
     }
-    =
-    \int_{0}^{\phi}
-    \frac{
-        dt
-    }{
-        \sqrt{
-            (1-t^2)(1-k^2 t^2)
-        }
-    } %TODO which is right? are both functions from phi?
 \end{equation}
-mit dem Winkel $\phi$ in Verbindung liegt.
+mit dem Winkel $\phi$ in Verbindung gebracht werden.
 
 Dabei wird das vollständige und unvollständige Elliptische integral unterschieden.
 Beim vollständigen Integral
@@ -53,9 +44,9 @@ Beim vollständigen Integral
         }
     }
 \end{equation}
-wird über ein viertel Ellipsenbogen integriert also bis $\phi=\pi/2$ und liefert das Winkelargument für eine Vierteldrehung.
+wird über ein viertel Ellipsenbogen integriert, also bis $\phi=\pi/2$ und liefert das Winkelargument für eine Vierteldrehung.
 Die Zahl wird oft auch abgekürzt mit $K = K(k)$ und ist für das elliptische Filter sehr relevant.
-Alle elliptishen Funktionen sind somit $4K$-periodisch.
+Alle elliptischen Funktionen sind somit $4K$-periodisch.
 
 Neben dem $\sn$ gibt es zwei weitere basis-elliptische Funktionen $\cn$ und $\dn$.
 Dazu kommen noch weitere abgeleitete Funktionen, die durch Quotienten und Kehrwerte dieser Funktionen zustande kommen.
@@ -93,37 +84,40 @@ Mithilfe von $F^{-1}$ kann zum Beispiel $sn^{-1}$ mit dem Elliptischen integral
     =
     \sn(z, k)
     =
-    w
-\end{equation}
-
-\begin{equation}
-    \phi
-    =
-     F^{-1}(z, k)
-     =
-     \sin^{-1} \big( \sn (z, k ) \big)
-     =
-    \sin^{-1} ( w )
-\end{equation}
-
-\begin{equation}
-    F(\phi, k)
-    =
-    z
-    =
-    F( \sin^{-1} \big( \sn (z, k ) \big) , k)
-    =
-    F( \sin^{-1} ( w ), k)
-\end{equation}
-
-\begin{equation}
-    \sn^{-1}(w, k)
-    =
-    F(\phi, k),
-    \quad
-    \phi = \sin^{-1}(w)
+    w.
 \end{equation}
 
+% \begin{equation} %TODO remove unnecessary equations
+%     \phi
+%     =
+%      F^{-1}(z, k)
+%      =
+%      \sin^{-1} \big( \sn (z, k ) \big)
+%      =
+%     \sin^{-1} ( w )
+% \end{equation}
+
+% \begin{equation}
+%     F(\phi, k)
+%     =
+%     z
+%     =
+%     F( \sin^{-1} \big( \sn (z, k ) \big) , k)
+%     =
+%     F( \sin^{-1} ( w ), k)
+% \end{equation}
+
+% \begin{equation}
+%     \sn^{-1}(w, k)
+%     =
+%     F(\phi, k),
+%     \quad
+%     \phi = \sin^{-1}(w)
+% \end{equation}
+
+Beim Tschebyscheff-Filter konnten wir mit Betrachten des Arcuscosinus die Funktionalität erklären.
+Für das Elliptische Filter machen wir die gleiche Betrachtung mit der $\sn^{-1}$-Funktion.
+Der $\sn^{-1}$ ist durch das elliptische Integral
 \begin{align}
     \sn^{-1}(w, k)
         & =
@@ -148,11 +142,8 @@ Mithilfe von $F^{-1}$ kann zum Beispiel $sn^{-1}$ mit dem Elliptischen integral
         }
     }
 \end{align}
-
-Beim $\cos^{-1}(x)$ haben wir gesehen, dass die analytische Fortsetzung bei $x < -1$ und $x > 1$ rechtwinklig in die Komplexen zahlen wandert.
-Wenn man das gleiche mit $\sn^{-1}(w, k)$ macht, erkennt man zwei interessante Stellen.
-Die erste ist die gleiche wie beim $\cos^{-1}(x)$ nämlich bei $t = \pm 1$.
-Der erste Term unter der Wurzel wird dann negativ, während der zweite noch positiv ist, da $k \leq 1$.
+beschrieben.
+Dazu betrachten wir wieder den Integranden
 \begin{equation}
     \frac{
         1
@@ -160,24 +151,15 @@ Der erste Term unter der Wurzel wird dann negativ, während der zweite noch posi
         \sqrt{
             (1-t^2)(1-k^2 t^2)
         }
-    }
-    \in \mathbb{R}
-    \quad \forall \quad
-    -1 \leq t \leq 1
+    }.
 \end{equation}
-Die zweite stelle passiert wenn beide Faktoren unter der Wurzel negativ werden, was bei $t = 1/k$ der Fall ist.
-
-
-
-
-Funktion in relle und komplexe Richtung periodisch
-
-In der reellen Richtung ist sie $4K(k)$-periodisch und in der imaginären Richtung $4K^\prime(k)$-periodisch.
-
-
-
-%TODO sn^{-1} grafik
-
+Beim $\cos^{-1}(x)$ haben wir gesehen, dass die analytische Fortsetzung bei $x < -1$ und $x > 1$ rechtwinklig in die Komplexen zahlen wandert.
+Wenn man das Gleiche mit $\sn^{-1}(w, k)$ macht, erkennt man zwei interessante Stellen.
+Die erste ist die gleiche wie beim $\cos^{-1}(x)$ nämlich bei $t = \pm 1$.
+Der erste Term unter der Wurzel wird dann negativ, während der zweite noch positiv ist, da $k \leq 1$.
+Ab diesem Punkt knickt die Funktion in die imaginäre Richtung ab.
+Bei $t = 1/k$ ist auch der zweite Term negativ und die Funktion verläuft in die negative reelle Richtung.
+Abbildung \ref{ellfilter:fig:sn} zeigt den Verlauf der Funktion in der komplexen Ebene.
 \begin{figure}
     \centering
     \input{papers/ellfilter/tikz/sn.tikz.tex}
@@ -185,5 +167,20 @@ In der reellen Richtung ist sie $4K(k)$-periodisch und in der imaginären Richtu
         $z$-Ebene der Funktion $z = \sn^{-1}(w, k)$.
         Die Funktion ist in der realen Achse $4K$-periodisch und in der imaginären Achse $2jK^\prime$-periodisch.
     }
-    % \label{ellfilter:fig:cd2}
+    \label{ellfilter:fig:sn}
 \end{figure}
+In der reellen Richtung ist sie $4K(k)$-periodisch und in der imaginären Richtung $4K^\prime(k)$-periodisch, wobei $K^\prime$ das komplementäre vollständige Elliptische Integral ist:
+\begin{equation}
+    K^\prime(k)
+    =
+    \int_{0}^{\pi / 2}
+    \frac{
+        d\theta
+    }{
+        \sqrt{
+            1-{k^\prime}^2 \sin^2 \theta
+        }
+    },
+    \quad
+    k^\prime = \sqrt{1-k^2}.
+\end{equation}
diff --git a/buch/papers/ellfilter/presentation/presentation.tex b/buch/papers/ellfilter/presentation/presentation.tex
index 7fdb864..96bdfd3 100644
--- a/buch/papers/ellfilter/presentation/presentation.tex
+++ b/buch/papers/ellfilter/presentation/presentation.tex
@@ -76,9 +76,9 @@
 
 %Title Page
 \title{Elliptische Filter}
-\subtitle{Eine Anwendung der Jaccobi elliptischen Funktionen}
+\subtitle{Eine Anwendung der Jacobi elliptischen Funktionen}
 \author{Nicolas Tobler}
-% \institute{OST Ostschweizer Fachhochschule}
+\institute{Mathematisches Seminar 2022 | Spezielle Funktionen}
 % \institute{\includegraphics[scale=0.3]{../img/ost_logo.png}}
 \date{\today}
 
@@ -113,25 +113,38 @@
 	\end{frame}
 
 	\begin{frame}
-		\frametitle{Content}
+		\frametitle{Inhalt}
 		\tableofcontents
 	\end{frame}
 
-	\section{Linear Filter}
+	\section{Lineare Filter}
 
 	\begin{frame}
 		\frametitle{Lineare Filter}
 
+		\begin{center}
+			\scalebox{0.75}{
+				\input{../tikz/filter.tikz.tex}
+			}
+		\end{center}
 
-		\begin{equation}
+
+		\begin{equation*}
 			| H(\Omega)|^2 = \frac{1}{1 + \varepsilon_p^2 F_N^2(w)}, \quad w=\frac{\Omega}{\Omega_p}
-		\end{equation}
+		\end{equation*}
 
 		\pause
 
-		\begin{equation}
+		\begin{align*}
+			|F_N(w)| &< 1 \quad \forall \quad |w| < 1 \\
+			|F_N(w)| &= 1 \quad \forall \quad |w| = 1 \\
+			|F_N(w)| &> 1 \quad \forall \quad |w| > 1
+		\end{align*}
+
+
+		\begin{equation*}
 			F_N(w) = w^N
-		\end{equation}
+		\end{equation*}
 
 	\end{frame}
 
@@ -218,10 +231,36 @@
 
 		Darstellung mit trigonometrischen Funktionen:
 
-		\begin{align} \label{ellfilter:eq:chebychef_polynomials}
+		\begin{align*}
 			T_N(w) &= \cos \left( N \cos^{-1}(w) \right) \\
 				&= \cos \left(N~z \right), \quad w= \cos(z)
-		\end{align}
+		\end{align*}
+
+		\pause
+
+		\begin{align*}
+			\cos^{-1}(x)
+			&=
+			\int_{x}^{1}
+			\frac{
+				dz
+			}{
+				\sqrt{
+					1-z^2
+				}
+			}\\
+			&=
+			\int_{0}^{x}
+			\frac{
+				-1
+			}{
+				\sqrt{
+					1-z^2
+				}
+			}
+			~dz
+			+ \frac{\pi}{2}
+		\end{align*}
 
 
 	\end{frame}
@@ -229,15 +268,41 @@
 	\begin{frame}
 		\frametitle{Tschebyscheff-Filter}
 
-		\begin{equation*}
-			z = \cos^{-1}(w)
-		\end{equation*}
+		\begin{columns}
+
+			\begin{column}{0.2\textwidth}
+
+				\begin{equation*}
+					z = \cos^{-1}(w)
+				\end{equation*}
+
+				\vspace{0.5cm}
+
+				Integrand:
+				\begin{equation*}
+					\frac{
+						-1
+					}{
+						\sqrt{
+							1-z^2
+						}
+					}
+				\end{equation*}
+
+			\end{column}
+			\begin{column}{0.8\textwidth}
+
+
+				\begin{center}
+					\scalebox{0.7}{
+						\input{../tikz/arccos.tikz.tex}
+					}
+				\end{center}
+
+			\end{column}
+		\end{columns}
+
 
-		\begin{center}
-			\scalebox{0.85}{
-				\input{../tikz/arccos.tikz.tex}
-			}
-		\end{center}
 
 	\end{frame}
 
@@ -245,7 +310,7 @@
 		\frametitle{Tschebyscheff-Filter}
 
 		\begin{equation*}
-			z_1 = N~\cos^{-1}(w)
+			T_N(w) = \cos \left(z_1 \right), \quad z_1 = N~\cos^{-1}(w)
 		\end{equation*}
 
 		\begin{center}
@@ -257,15 +322,14 @@
 	\end{frame}
 
 
-	\section{Jaccobi elliptische Funktionen}
+	\section{Jacobi elliptische Funktionen}
 
 	\begin{frame}
-		\frametitle{Jaccobi elliptische Funktionen}
+		\frametitle{Jacobi elliptische Funktionen}
 
+		Elliptisches Integral erster Art
 
-		\begin{equation}
-			z
-			=
+		\begin{equation*}
 			F(\phi, k)
 			=
 			\int_{0}^{\phi}
@@ -276,18 +340,18 @@
 					1-k^2 \sin^2 \theta
 				}
 			}
-			=
-			\int_{0}^{\phi}
-			\frac{
-				dt
-			}{
-				\sqrt{
-					(1-t^2)(1-k^2 t^2)
-				}
-			}
-		\end{equation}
+			% =
+			% \int_{0}^{\phi}
+			% \frac{
+			% 	dt
+			% }{
+			% 	\sqrt{
+			% 		(1-t^2)(1-k^2 t^2)
+			% 	}
+			% }
+		\end{equation*}
 
-		\begin{equation}
+		\begin{equation*}
 			K(k)
 			=
 			\int_{0}^{\pi / 2}
@@ -298,24 +362,88 @@
 					1-k^2 \sin^2 \theta
 				}
 			}
-		\end{equation}
+		\end{equation*}
 
 
 
 	\end{frame}
 
+
+
+
+
 	\begin{frame}
-		\frametitle{Jaccobi elliptische Funktionen}
+		\frametitle{Jacobi elliptische Funktionen}
+
+			\begin{equation*}
+				\sn^{-1}(w, k)
+				=
+				F(\phi, k),
+				\quad
+				\phi = \sin^{-1}(w)
+			\end{equation*}
+
+			\begin{align*}
+				\sn^{-1}(w, k)
+					& =
+				\int_{0}^{\phi}
+				\frac{
+					d\theta
+				}{
+					\sqrt{
+						1-k^2 \sin^2 \theta
+					}
+				},
+				\quad
+				\phi = \sin^{-1}(w)
+				\\
+					& =
+				\int_{0}^{w}
+				\frac{
+					dt
+				}{
+					\sqrt{
+						(1-t^2)(1-k^2 t^2)
+					}
+				}
+			\end{align*}
 
-		\begin{equation*}
-			z = \sn^{-1}(w, k)
-		\end{equation*}
 
-		\begin{center}
-			\scalebox{0.7}{
-				\input{../tikz/sn.tikz.tex}
-			}
-		\end{center}
+
+		\end{frame}
+
+	\begin{frame}
+		\frametitle{Jacobi elliptische Funktionen}
+		\begin{columns}
+			\begin{column}{0.2\textwidth}
+
+				\begin{equation*}
+					z = \sn^{-1}(w, k)
+				\end{equation*}
+
+				\vspace{0.5cm}
+
+				Integrand:
+				\begin{equation*}
+					\frac{
+						1
+					}{
+						\sqrt{
+							(1-t^2)(1-k^2 t^2)
+						}
+					}
+				\end{equation*}
+
+			\end{column}
+			\begin{column}{0.8\textwidth}
+				\begin{center}
+					\scalebox{0.75}{
+						\input{../tikz/sn.tikz.tex}
+					}
+				\end{center}
+			\end{column}
+		\end{columns}
+
 
 	\end{frame}
 
@@ -334,7 +462,7 @@
 
 
 	\begin{frame}
-		\frametitle{Jaccobi elliptische Funktionen}
+		\frametitle{Jacobi elliptische Funktionen}
 
 		\begin{equation*}
 			z = \cd^{-1}(w, k)
@@ -349,6 +477,23 @@
 
 	\end{frame}
 
+	\section{Elliptisches Filter}
+
+	\begin{frame}
+		\frametitle{Elliptisches Filter}
+
+		% \begin{equation*}
+		% 	z_1 = N~\frac{K_1}{K}~\cd^{-1}(w, k)
+		% \end{equation*}
+
+		\begin{center}
+			\scalebox{0.75}{
+				\input{../tikz/cd3.tikz.tex}
+			}
+		\end{center}
+
+	\end{frame}
+
 	\begin{frame}
 		\frametitle{Periodizität in realer und imaginärer Richtung}
 
@@ -360,20 +505,42 @@
 	\end{frame}
 
 	\begin{frame}
+		\frametitle{Gradgleichung}
+
+		\begin{center}
+			\scalebox{0.95}{
+				\input{../tikz/elliptic_transform2.tikz}
+			}
+		\end{center}
+
+		\onslide<5->{
+		\begin{equation*}
+			N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1}
+		\end{equation*}
+		}
+
+	\end{frame}
+
+	\begin{frame}
 		\frametitle{Elliptisches Filter}
 
 		\begin{equation*}
-			z_1 = N~\frac{K_1}{K}~\cd^{-1}(w, k)
+			R_N = \cd(z_1, k_1),
+			\quad
+			z_1 = N~\frac{K_1}{K}~\cd^{-1}(w, k),
+			\quad
+			N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1}
 		\end{equation*}
 
 		\begin{center}
-			\scalebox{0.8}{
+			\scalebox{0.75}{
 				\input{../tikz/cd2.tikz.tex}
 			}
 		\end{center}
 
 	\end{frame}
 
+
 	\begin{frame}
 		\frametitle{Elliptisches Filter}
 
@@ -401,13 +568,4 @@
 
 	\end{frame}
 
-	\begin{frame}
-		\frametitle{Gradgleichung}
-
-		\begin{equation}
-			N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1}
-		\end{equation}
-
-	\end{frame}
-
 	\end{document}
diff --git a/buch/papers/ellfilter/python/F_N_elliptic.pgf b/buch/papers/ellfilter/python/F_N_elliptic.pgf
deleted file mode 100644
index 03084c6..0000000
--- a/buch/papers/ellfilter/python/F_N_elliptic.pgf
+++ /dev/null
@@ -1,847 +0,0 @@
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diff --git a/buch/papers/ellfilter/python/elliptic.pgf b/buch/papers/ellfilter/python/elliptic.pgf
index 31b77d4..32485c1 100644
--- a/buch/papers/ellfilter/python/elliptic.pgf
+++ b/buch/papers/ellfilter/python/elliptic.pgf
@@ -23,7 +23,7 @@
 \begingroup%
 \makeatletter%
 \begin{pgfpicture}%
-\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{4.000000in}{2.500000in}}%
+\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{5.000000in}{3.000000in}}%
 \pgfusepath{use as bounding box, clip}%
 \begin{pgfscope}%
 \pgfsetbuttcap%
@@ -34,9 +34,9 @@
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 \pgfsetdash{}{0pt}%
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-\pgfpathlineto{\pgfqpoint{4.000000in}{0.000000in}}%
-\pgfpathlineto{\pgfqpoint{4.000000in}{2.500000in}}%
-\pgfpathlineto{\pgfqpoint{0.000000in}{2.500000in}}%
+\pgfpathlineto{\pgfqpoint{5.000000in}{0.000000in}}%
+\pgfpathlineto{\pgfqpoint{5.000000in}{3.000000in}}%
+\pgfpathlineto{\pgfqpoint{0.000000in}{3.000000in}}%
 \pgfpathlineto{\pgfqpoint{0.000000in}{0.000000in}}%
 \pgfpathclose%
 \pgfusepath{}%
@@ -51,16 +51,16 @@
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-\pgfpathlineto{\pgfqpoint{0.617954in}{0.548769in}}%
+\pgfpathmoveto{\pgfqpoint{0.733531in}{1.746607in}}%
+\pgfpathlineto{\pgfqpoint{4.727004in}{1.746607in}}%
+\pgfpathlineto{\pgfqpoint{4.727004in}{2.850000in}}%
+\pgfpathlineto{\pgfqpoint{0.733531in}{2.850000in}}%
+\pgfpathlineto{\pgfqpoint{0.733531in}{1.746607in}}%
 \pgfpathclose%
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 \end{pgfscope}%
 \begin{pgfscope}%
-\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}%
+\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}%
 \pgfusepath{clip}%
 \pgfsetbuttcap%
 \pgfsetmiterjoin%
@@ -72,16 +72,16 @@
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-\pgfpathmoveto{\pgfqpoint{0.617954in}{1.788459in}}%
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-\pgfpathlineto{\pgfqpoint{0.617954in}{3.541645in}}%
-\pgfpathlineto{\pgfqpoint{0.617954in}{1.788459in}}%
+\pgfpathmoveto{\pgfqpoint{0.733531in}{-108.151374in}}%
+\pgfpathlineto{\pgfqpoint{2.730268in}{-108.151374in}}%
+\pgfpathlineto{\pgfqpoint{2.730268in}{2.187964in}}%
+\pgfpathlineto{\pgfqpoint{0.733531in}{2.187964in}}%
+\pgfpathlineto{\pgfqpoint{0.733531in}{-108.151374in}}%
 \pgfpathclose%
 \pgfusepath{fill}%
 \end{pgfscope}%
 \begin{pgfscope}%
-\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}%
+\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}%
 \pgfusepath{clip}%
 \pgfsetbuttcap%
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@@ -93,16 +93,16 @@
 \pgfsetstrokecolor{currentstroke}%
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-\pgfpathmoveto{\pgfqpoint{2.189776in}{0.724087in}}%
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-\pgfpathlineto{\pgfqpoint{2.205494in}{1.788459in}}%
-\pgfpathlineto{\pgfqpoint{2.189776in}{1.788459in}}%
-\pgfpathlineto{\pgfqpoint{2.189776in}{0.724087in}}%
+\pgfpathmoveto{\pgfqpoint{2.730268in}{2.187964in}}%
+\pgfpathlineto{\pgfqpoint{2.750075in}{2.187964in}}%
+\pgfpathlineto{\pgfqpoint{2.750075in}{2.408643in}}%
+\pgfpathlineto{\pgfqpoint{2.730268in}{2.408643in}}%
+\pgfpathlineto{\pgfqpoint{2.730268in}{2.187964in}}%
 \pgfpathclose%
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 \begin{pgfscope}%
-\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}%
+\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}%
 \pgfusepath{clip}%
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@@ -114,16 +114,16 @@
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 \pgfpathclose%
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@@ -131,8 +131,8 @@
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 \pgfusepath{stroke}%
 \end{pgfscope}%
 \begin{pgfscope}%
@@ -150,7 +150,295 @@
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+\pgfsys@transformshift{0.733531in}{1.746607in}%
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+}%
+\begin{pgfscope}%
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+}%
+\begin{pgfscope}%
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+}%
+\begin{pgfscope}%
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+\pgftext[x=0.733531in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.00}\)}%
+\end{pgfscope}%
+\begin{pgfscope}%
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+\pgfpathlineto{\pgfqpoint{1.232715in}{1.473611in}}%
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 \end{pgfscope}%
 \begin{pgfscope}%
@@ -188,7 +951,7 @@
 \pgfusepath{stroke,fill}%
 }%
 \begin{pgfscope}%
-\pgfsys@transformshift{1.403865in}{0.548769in}%
+\pgfsys@transformshift{1.232715in}{0.370218in}%
 \pgfsys@useobject{currentmarker}{}%
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 \end{pgfscope}%
@@ -196,10 +959,10 @@
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-\pgftext[x=1.403865in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.5}\)}%
+\pgftext[x=1.232715in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.25}\)}%
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 \begin{pgfscope}%
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+\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}%
 \pgfusepath{clip}%
 \pgfsetrectcap%
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@@ -207,8 +970,8 @@
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-\pgfpathmoveto{\pgfqpoint{2.189776in}{0.548769in}}%
-\pgfpathlineto{\pgfqpoint{2.189776in}{2.301955in}}%
+\pgfpathmoveto{\pgfqpoint{1.731899in}{0.370218in}}%
+\pgfpathlineto{\pgfqpoint{1.731899in}{1.473611in}}%
 \pgfusepath{stroke}%
 \end{pgfscope}%
 \begin{pgfscope}%
@@ -226,7 +989,7 @@
 \pgfusepath{stroke,fill}%
 }%
 \begin{pgfscope}%
-\pgfsys@transformshift{2.189776in}{0.548769in}%
+\pgfsys@transformshift{1.731899in}{0.370218in}%
 \pgfsys@useobject{currentmarker}{}%
 \end{pgfscope}%
 \end{pgfscope}%
@@ -234,10 +997,10 @@
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-\pgftext[x=2.189776in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.0}\)}%
+\pgftext[x=1.731899in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.50}\)}%
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 \begin{pgfscope}%
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 \pgfusepath{clip}%
 \pgfsetrectcap%
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@@ -245,8 +1008,8 @@
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-\pgfpathlineto{\pgfqpoint{2.975686in}{2.301955in}}%
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 \pgfusepath{stroke}%
 \end{pgfscope}%
 \begin{pgfscope}%
@@ -264,7 +1027,7 @@
 \pgfusepath{stroke,fill}%
 }%
 \begin{pgfscope}%
-\pgfsys@transformshift{2.975686in}{0.548769in}%
+\pgfsys@transformshift{2.231083in}{0.370218in}%
 \pgfsys@useobject{currentmarker}{}%
 \end{pgfscope}%
 \end{pgfscope}%
@@ -272,10 +1035,10 @@
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-\pgftext[x=2.975686in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.5}\)}%
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 \end{pgfscope}%
 \begin{pgfscope}%
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+\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}%
 \pgfusepath{clip}%
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@@ -283,8 +1046,8 @@
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-\pgfpathlineto{\pgfqpoint{3.761597in}{2.301955in}}%
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 \pgfusepath{stroke}%
 \end{pgfscope}%
 \begin{pgfscope}%
@@ -302,7 +1065,7 @@
 \pgfusepath{stroke,fill}%
 }%
 \begin{pgfscope}%
-\pgfsys@transformshift{3.761597in}{0.548769in}%
+\pgfsys@transformshift{2.730268in}{0.370218in}%
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 \end{pgfscope}%
 \end{pgfscope}%
@@ -310,16 +1073,48 @@
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+}%
+\begin{pgfscope}%
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-\pgfpathlineto{\pgfqpoint{3.761597in}{0.717600in}}%
+\definecolor{currentstroke}{rgb}{1.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{currentstroke}%
+\pgfsetdash{}{0pt}%
+\pgfpathmoveto{\pgfqpoint{2.750075in}{0.480388in}}%
+\pgfpathlineto{\pgfqpoint{2.756998in}{0.398490in}}%
+\pgfpathlineto{\pgfqpoint{2.759965in}{0.373708in}}%
+\pgfpathlineto{\pgfqpoint{2.760954in}{0.373969in}}%
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+\pgfpathlineto{\pgfqpoint{2.886552in}{0.448435in}}%
+\pgfpathlineto{\pgfqpoint{2.959734in}{0.406057in}}%
+\pgfpathlineto{\pgfqpoint{3.006216in}{0.383080in}}%
+\pgfpathlineto{\pgfqpoint{3.036873in}{0.370620in}}%
+\pgfpathlineto{\pgfqpoint{3.085332in}{0.388800in}}%
+\pgfpathlineto{\pgfqpoint{3.137747in}{0.405140in}}%
+\pgfpathlineto{\pgfqpoint{3.195107in}{0.419767in}}%
+\pgfpathlineto{\pgfqpoint{3.259389in}{0.432926in}}%
+\pgfpathlineto{\pgfqpoint{3.332572in}{0.444663in}}%
+\pgfpathlineto{\pgfqpoint{3.415645in}{0.454798in}}%
+\pgfpathlineto{\pgfqpoint{3.512563in}{0.463443in}}%
+\pgfpathlineto{\pgfqpoint{3.627282in}{0.470496in}}%
+\pgfpathlineto{\pgfqpoint{3.766725in}{0.475859in}}%
+\pgfpathlineto{\pgfqpoint{3.940782in}{0.479318in}}%
+\pgfpathlineto{\pgfqpoint{4.167253in}{0.480557in}}%
+\pgfpathlineto{\pgfqpoint{4.484709in}{0.478977in}}%
+\pgfpathlineto{\pgfqpoint{4.727004in}{0.476474in}}%
+\pgfpathlineto{\pgfqpoint{4.727004in}{0.476474in}}%
 \pgfusepath{stroke}%
 \end{pgfscope}%
 \begin{pgfscope}%
@@ -667,8 +1482,8 @@
 \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
 \pgfsetstrokecolor{currentstroke}%
 \pgfsetdash{}{0pt}%
-\pgfpathmoveto{\pgfqpoint{0.617954in}{0.548769in}}%
-\pgfpathlineto{\pgfqpoint{0.617954in}{2.301955in}}%
+\pgfpathmoveto{\pgfqpoint{0.733531in}{0.370218in}}%
+\pgfpathlineto{\pgfqpoint{0.733531in}{1.473611in}}%
 \pgfusepath{stroke}%
 \end{pgfscope}%
 \begin{pgfscope}%
@@ -678,8 +1493,8 @@
 \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
 \pgfsetstrokecolor{currentstroke}%
 \pgfsetdash{}{0pt}%
-\pgfpathmoveto{\pgfqpoint{3.761597in}{0.548769in}}%
-\pgfpathlineto{\pgfqpoint{3.761597in}{2.301955in}}%
+\pgfpathmoveto{\pgfqpoint{4.727004in}{0.370218in}}%
+\pgfpathlineto{\pgfqpoint{4.727004in}{1.473611in}}%
 \pgfusepath{stroke}%
 \end{pgfscope}%
 \begin{pgfscope}%
@@ -689,8 +1504,8 @@
 \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
 \pgfsetstrokecolor{currentstroke}%
 \pgfsetdash{}{0pt}%
-\pgfpathmoveto{\pgfqpoint{0.617954in}{0.548769in}}%
-\pgfpathlineto{\pgfqpoint{3.761597in}{0.548769in}}%
+\pgfpathmoveto{\pgfqpoint{0.733531in}{0.370218in}}%
+\pgfpathlineto{\pgfqpoint{4.727004in}{0.370218in}}%
 \pgfusepath{stroke}%
 \end{pgfscope}%
 \begin{pgfscope}%
@@ -700,8 +1515,8 @@
 \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
 \pgfsetstrokecolor{currentstroke}%
 \pgfsetdash{}{0pt}%
-\pgfpathmoveto{\pgfqpoint{0.617954in}{2.301955in}}%
-\pgfpathlineto{\pgfqpoint{3.761597in}{2.301955in}}%
+\pgfpathmoveto{\pgfqpoint{0.733531in}{1.473611in}}%
+\pgfpathlineto{\pgfqpoint{4.727004in}{1.473611in}}%
 \pgfusepath{stroke}%
 \end{pgfscope}%
 \end{pgfpicture}%
diff --git a/buch/papers/ellfilter/python/elliptic.py b/buch/papers/ellfilter/python/elliptic.py
index b3336a1..6e0fd12 100644
--- a/buch/papers/ellfilter/python/elliptic.py
+++ b/buch/papers/ellfilter/python/elliptic.py
@@ -324,9 +324,9 @@ K_prime = ell_int(np.sqrt(1-k**2))
 
 
 f, axs = plt.subplots(1,2, figsize=(5,2.5))
-axs[0].plot(k, K, linewidth=0.1)
+axs[0].plot(k, K, linewidth=1)
 axs[0].text(k[30], K[30]+0.1, f"$K$")
-axs[0].plot(k, K_prime, linewidth=0.1)
+axs[0].plot(k, K_prime, linewidth=1)
 axs[0].text(k[30], K_prime[30]+0.1, f"$K^\prime$")
 axs[0].set_xlim([0,1])
 axs[0].set_ylim([0,4])
@@ -342,9 +342,9 @@ k = np.array([0.1,0.2,0.4,0.6,0.9,0.99])
 K = ell_int(k)
 K_prime = ell_int(np.sqrt(1-k**2))
 
-axs[1].plot(K, K_prime, '.', color=last_color(), markersize=2)
-for x, y, n in zip(K, K_prime, k):
-    axs[1].text(x+0.1, y+0.1, f"$k={n:.2f}$", rotation_mode="anchor")
+# axs[1].plot(K, K_prime, '.', color=last_color(), markersize=2)
+# for x, y, n in zip(K, K_prime, k):
+#     axs[1].text(x+0.1, y+0.1, f"$k={n:.2f}$", rotation_mode="anchor")
 axs[1].set_ylabel("$K^\prime$")
 axs[1].set_xlabel("$K$")
 axs[1].set_xlim([0,6])
diff --git a/buch/papers/ellfilter/python/elliptic2.py b/buch/papers/ellfilter/python/elliptic2.py
index 29c6f47..20a7428 100644
--- a/buch/papers/ellfilter/python/elliptic2.py
+++ b/buch/papers/ellfilter/python/elliptic2.py
@@ -1,5 +1,6 @@
 # %%
 
+import enum
 import matplotlib.pyplot as plt
 import scipy.signal
 import numpy as np
@@ -8,7 +9,9 @@ from matplotlib.patches import Rectangle
 
 import plot_params
 
-def ellip_filter(N):
+N=5
+
+def ellip_filter(N, mode=-1):
 
     order = N
     passband_ripple_db = 3
@@ -26,7 +29,16 @@ def ellip_filter(N):
         fs=None
     )
 
-    w, mag_db, phase = scipy.signal.bode((a, b), w=np.linspace(0*omega_c,2*omega_c, 4000))
+    if mode == 0:
+        w = np.linspace(0*omega_c,omega_c, 2000)
+    elif mode == 1:
+        w = np.linspace(omega_c,1.00992*omega_c, 2000)
+    elif mode == 2:
+        w = np.linspace(1.00992*omega_c,2*omega_c, 2000)
+    else:
+        w = np.linspace(0*omega_c,2*omega_c, 4000)
+
+    w, mag_db, phase = scipy.signal.bode((a, b), w=w)
 
     mag = 10**(mag_db/20)
 
@@ -38,105 +50,96 @@ def ellip_filter(N):
     return w/omega_c, FN2 / epsilon2, mag, a, b
 
 
-plt.figure(figsize=(4,2.5))
+f, axs = plt.subplots(2, 1, figsize=(5,3), sharex=True)
 
-for N in [5]:
-    w, FN2, mag, a, b = ellip_filter(N)
-    plt.semilogy(w, FN2, label=f"$N={N}, k=0.1$", linewidth=1)
+for mode, c in enumerate(["green", "orange", "red"]):
+    w, FN2, mag, a, b = ellip_filter(N, mode=mode)
+    axs[0].semilogy(w, FN2, label=f"$N={N}, k=0.1$", linewidth=1, color=c)
 
-plt.gca().add_patch(Rectangle(
+axs[0].add_patch(Rectangle(
     (0, 0),
     1, 1,
     fc ='green',
     alpha=0.2,
     lw = 10,
 ))
-plt.gca().add_patch(Rectangle(
+axs[0].add_patch(Rectangle(
     (1, 1),
-    0.01, 1e2-1,
+    0.00992, 1e2-1,
     fc ='orange',
     alpha=0.2,
     lw = 10,
 ))
 
-plt.gca().add_patch(Rectangle(
-    (1.01, 100),
+axs[0].add_patch(Rectangle(
+    (1.00992, 100),
     1, 1e6,
     fc ='red',
     alpha=0.2,
     lw = 10,
 ))
 
-zeros = [0,0.87,1]
+zeros = [0,0.87,0.995]
 poles = [1.01,1.155]
 
 import matplotlib.transforms
-plt.plot( # mark errors as vertical bars
+axs[0].plot( # mark errors as vertical bars
     zeros,
     np.zeros_like(zeros),
     "o",
     mfc='none',
     color='black',
     transform=matplotlib.transforms.blended_transform_factory(
-        plt.gca().transData,
-        plt.gca().transAxes,
+        axs[0].transData,
+        axs[0].transAxes,
     ),
 )
-plt.plot( # mark errors as vertical bars
+axs[0].plot( # mark errors as vertical bars
     poles,
     np.ones_like(poles),
     "x",
     mfc='none',
     color='black',
     transform=matplotlib.transforms.blended_transform_factory(
-        plt.gca().transData,
-        plt.gca().transAxes,
+        axs[0].transData,
+        axs[0].transAxes,
     ),
 )
 
-plt.xlim([0,2])
-plt.ylim([1e-4,1e6])
-plt.grid()
-plt.xlabel("$w$")
-plt.ylabel("$F^2_N(w)$")
-plt.legend()
-plt.tight_layout()
-plt.savefig("F_N_elliptic.pgf")
-plt.show()
-
-
-
-plt.figure(figsize=(4,2.5))
-plt.plot(w, mag, linewidth=1)
+for mode, c in enumerate(["green", "orange", "red"]):
+    w, FN2, mag, a, b = ellip_filter(N, mode=mode)
+    axs[1].plot(w, mag, linewidth=1, color=c)
 
-plt.gca().add_patch(Rectangle(
+axs[1].add_patch(Rectangle(
     (0, np.sqrt(2)/2),
     1, 1,
     fc ='green',
     alpha=0.2,
     lw = 10,
 ))
-plt.gca().add_patch(Rectangle(
+axs[1].add_patch(Rectangle(
     (1, 0.1),
-    0.01, np.sqrt(2)/2 - 0.1,
+    0.00992, np.sqrt(2)/2 - 0.1,
     fc ='orange',
     alpha=0.2,
     lw = 10,
 ))
 
-plt.gca().add_patch(Rectangle(
-    (1.01, 0),
+axs[1].add_patch(Rectangle(
+    (1.00992, 0),
     1, 0.1,
     fc ='red',
     alpha=0.2,
     lw = 10,
 ))
 
-plt.grid()
-plt.xlim([0,2])
-plt.ylim([0,1])
-plt.xlabel("$w$")
-plt.ylabel("$|H(w)|$")
+axs[0].set_xlim([0,2])
+axs[0].set_ylim([1e-4,1e6])
+axs[0].grid()
+axs[0].set_ylabel("$F^2_N(w)$")
+axs[1].grid()
+axs[1].set_ylim([0,1])
+axs[1].set_ylabel("$|H(w)|$")
 plt.tight_layout()
 plt.savefig("elliptic.pgf")
 plt.show()
diff --git a/buch/papers/ellfilter/python/k.pgf b/buch/papers/ellfilter/python/k.pgf
index 95d61d4..bbb823a 100644
--- a/buch/papers/ellfilter/python/k.pgf
+++ b/buch/papers/ellfilter/python/k.pgf
@@ -320,7 +320,7 @@
 \pgfusepath{clip}%
 \pgfsetrectcap%
 \pgfsetroundjoin%
-\pgfsetlinewidth{0.100375pt}%
+\pgfsetlinewidth{1.003750pt}%
 \definecolor{currentstroke}{rgb}{0.121569,0.466667,0.705882}%
 \pgfsetstrokecolor{currentstroke}%
 \pgfsetdash{}{0pt}%
@@ -434,7 +434,7 @@
 \pgfusepath{clip}%
 \pgfsetrectcap%
 \pgfsetroundjoin%
-\pgfsetlinewidth{0.100375pt}%
+\pgfsetlinewidth{1.003750pt}%
 \definecolor{currentstroke}{rgb}{1.000000,0.498039,0.054902}%
 \pgfsetstrokecolor{currentstroke}%
 \pgfsetdash{}{0pt}%
@@ -1011,56 +1011,6 @@
 \pgfusepath{stroke}%
 \end{pgfscope}%
 \begin{pgfscope}%
-\pgfpathrectangle{\pgfqpoint{2.874885in}{0.548769in}}{\pgfqpoint{1.940523in}{1.753186in}}%
-\pgfusepath{clip}%
-\pgfsetbuttcap%
-\pgfsetroundjoin%
-\definecolor{currentfill}{rgb}{0.121569,0.466667,0.705882}%
-\pgfsetfillcolor{currentfill}%
-\pgfsetlinewidth{1.003750pt}%
-\definecolor{currentstroke}{rgb}{0.121569,0.466667,0.705882}%
-\pgfsetstrokecolor{currentstroke}%
-\pgfsetdash{}{0pt}%
-\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.006944in}{-0.006944in}}{\pgfqpoint{0.006944in}{0.006944in}}{%
-\pgfpathmoveto{\pgfqpoint{0.000000in}{-0.006944in}}%
-\pgfpathcurveto{\pgfqpoint{0.001842in}{-0.006944in}}{\pgfqpoint{0.003608in}{-0.006213in}}{\pgfqpoint{0.004910in}{-0.004910in}}%
-\pgfpathcurveto{\pgfqpoint{0.006213in}{-0.003608in}}{\pgfqpoint{0.006944in}{-0.001842in}}{\pgfqpoint{0.006944in}{0.000000in}}%
-\pgfpathcurveto{\pgfqpoint{0.006944in}{0.001842in}}{\pgfqpoint{0.006213in}{0.003608in}}{\pgfqpoint{0.004910in}{0.004910in}}%
-\pgfpathcurveto{\pgfqpoint{0.003608in}{0.006213in}}{\pgfqpoint{0.001842in}{0.006944in}}{\pgfqpoint{0.000000in}{0.006944in}}%
-\pgfpathcurveto{\pgfqpoint{-0.001842in}{0.006944in}}{\pgfqpoint{-0.003608in}{0.006213in}}{\pgfqpoint{-0.004910in}{0.004910in}}%
-\pgfpathcurveto{\pgfqpoint{-0.006213in}{0.003608in}}{\pgfqpoint{-0.006944in}{0.001842in}}{\pgfqpoint{-0.006944in}{0.000000in}}%
-\pgfpathcurveto{\pgfqpoint{-0.006944in}{-0.001842in}}{\pgfqpoint{-0.006213in}{-0.003608in}}{\pgfqpoint{-0.004910in}{-0.004910in}}%
-\pgfpathcurveto{\pgfqpoint{-0.003608in}{-0.006213in}}{\pgfqpoint{-0.001842in}{-0.006944in}}{\pgfqpoint{0.000000in}{-0.006944in}}%
-\pgfpathlineto{\pgfqpoint{0.000000in}{-0.006944in}}%
-\pgfpathclose%
-\pgfusepath{stroke,fill}%
-}%
-\begin{pgfscope}%
-\pgfsys@transformshift{3.384190in}{1.844597in}%
-\pgfsys@useobject{currentmarker}{}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\pgfsys@transformshift{3.388110in}{1.606330in}%
-\pgfsys@useobject{currentmarker}{}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\pgfsys@transformshift{3.405294in}{1.376014in}%
-\pgfsys@useobject{currentmarker}{}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\pgfsys@transformshift{3.441114in}{1.248396in}%
-\pgfsys@useobject{currentmarker}{}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\pgfsys@transformshift{3.612461in}{1.128939in}%
-\pgfsys@useobject{currentmarker}{}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\pgfsys@transformshift{3.960478in}{1.102320in}%
-\pgfsys@useobject{currentmarker}{}%
-\end{pgfscope}%
-\end{pgfscope}%
-\begin{pgfscope}%
 \pgfsetrectcap%
 \pgfsetmiterjoin%
 \pgfsetlinewidth{0.803000pt}%
@@ -1116,42 +1066,6 @@
 \pgfsetfillcolor{textcolor}%
 \pgftext[x=3.415254in,y=0.583833in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle \pi/2\)}%
 \end{pgfscope}%
-\begin{pgfscope}%
-\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
-\pgfsetstrokecolor{textcolor}%
-\pgfsetfillcolor{textcolor}%
-\pgftext[x=3.416532in,y=1.879661in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.10\)}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
-\pgfsetstrokecolor{textcolor}%
-\pgfsetfillcolor{textcolor}%
-\pgftext[x=3.420452in,y=1.641394in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.20\)}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
-\pgfsetstrokecolor{textcolor}%
-\pgfsetfillcolor{textcolor}%
-\pgftext[x=3.437636in,y=1.411078in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.40\)}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
-\pgfsetstrokecolor{textcolor}%
-\pgfsetfillcolor{textcolor}%
-\pgftext[x=3.473456in,y=1.283460in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.60\)}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
-\pgfsetstrokecolor{textcolor}%
-\pgfsetfillcolor{textcolor}%
-\pgftext[x=3.644803in,y=1.164003in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.90\)}%
-\end{pgfscope}%
-\begin{pgfscope}%
-\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
-\pgfsetstrokecolor{textcolor}%
-\pgfsetfillcolor{textcolor}%
-\pgftext[x=3.992820in,y=1.137383in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.99\)}%
-\end{pgfscope}%
 \end{pgfpicture}%
 \makeatother%
 \endgroup%
diff --git a/buch/papers/ellfilter/tikz/arccos.tikz.tex b/buch/papers/ellfilter/tikz/arccos.tikz.tex
index 2772620..b11c25d 100644
--- a/buch/papers/ellfilter/tikz/arccos.tikz.tex
+++ b/buch/papers/ellfilter/tikz/arccos.tikz.tex
@@ -8,29 +8,6 @@
 
     \begin{scope}[xscale=0.6]
 
-        \clip(-7.5,-2) rectangle (7.5,2);
-
-        \draw[thick, ->, darkgreen] (0, 0) -- (0,1.5);
-        \draw[thick, ->, orange] (1, 0) -- (0,0);
-        \draw[thick, ->, red] (2, 0) -- (1,0);
-        \draw[thick, ->, blue] (2,1.5) -- (2, 0);
-
-        \foreach \i in {-2,...,1} {
-            \begin{scope}[opacity=0.5, xshift=\i*4cm]
-                \draw[->, orange] (-1, 0) -- (0,0);
-                \draw[->, darkgreen] (0, 0) -- (0,1.5);
-                \draw[->, darkgreen] (0, 0) -- (0,-1.5);
-                \draw[->, orange] (1, 0) -- (0,0);
-                \draw[->, red] (2, 0) -- (1,0);
-                \draw[->, blue] (2,1.5) -- (2, 0);
-                \draw[->, blue] (2,-1.5) -- (2, 0);
-                \draw[->, red] (2, 0) -- (3,0);
-
-                \node[zero] at (1,0) {};
-                \node[zero] at (3,0) {};
-            \end{scope}
-        }
-
         \node[gray, anchor=north] at (-6,0) {$-3\pi$};
         \node[gray, anchor=north] at (-4,0) {$-2\pi$};
         \node[gray, anchor=north] at (-2,0) {$-\pi$};
@@ -43,16 +20,53 @@
         % \node[gray, anchor=south east] at (0, 0) {$0$};
         \node[gray, anchor=east] at (0, 1.5) {$\infty$};
 
+        \clip(-7.5,-2) rectangle (7.5,2);
+
+        % \pause
+        \draw[ultra thick, ->, darkgreen] (1, 0) -- (0,0);
+        % \pause
+        \draw[ultra thick, ->, orange] (0, 0) -- (0,1.5);
+        % \pause
+        \draw[ultra thick, ->, cyan] (2, 0) -- (1,0);
+        \draw[ultra thick, ->, blue] (2,1.5) -- (2, 0);
+
+        % \pause
+
+        \foreach \i in {-2,...,1} {
+            \begin{scope}[xshift=\i*4cm]
+                \begin{scope}[]
+                    \draw[->, darkgreen] (-1, 0) -- (0,0);
+                    \draw[->, orange] (0, 0) -- (0,1.5);
+                    \draw[->, orange] (0, 0) -- (0,-1.5);
+                    \draw[->, darkgreen] (1, 0) -- (0,0);
+                    \draw[->, cyan] (2, 0) -- (1,0);
+                    \draw[->, blue] (2,1.5) -- (2, 0);
+                    \draw[->, blue] (2,-1.5) -- (2, 0);
+                    \draw[->, cyan] (2, 0) -- (3,0);
+                \end{scope}
+                \node[zero] at (1,0) {};
+                \node[zero] at (3,0) {};
+            \end{scope}
+        }
+
     \end{scope}
 
-    \begin{scope}[yshift=-2.5cm]
+    \node[zero] at (4,2) (n) {};
+    \node[anchor=west] at (n.east) {Nullstelle};
+
+    \begin{scope}[yshift=-3.25cm]
+
+        \draw[->, thick](0,0) -- node[anchor=center, fill=white]{$z = \cos^{-1}(w)$} (0,1);
+
+    \end{scope}
+    \begin{scope}[yshift=-4cm]
 
         \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$w$};
 
-        \draw[thick, ->, blue]      (-4, 0) -- (-2, 0);
-        \draw[thick, ->, red]       (-2, 0) -- (0, 0);
-        \draw[thick, ->, orange]    (0, 0) -- (2, 0);
-        \draw[thick, ->, darkgreen] (2, 0) -- (4, 0);
+        \draw[ultra thick, ->, blue]      (-4, 0) -- (-2, 0);
+        \draw[ultra thick, ->, cyan]       (-2, 0) -- (0, 0);
+        \draw[ultra thick, ->, darkgreen]    (0, 0) -- (2, 0);
+        \draw[ultra thick, ->, orange] (2, 0) -- (4, 0);
 
         \node[anchor=south] at (-4,0) {$-\infty$};
         \node[anchor=south] at (-2,0) {$-1$};
diff --git a/buch/papers/ellfilter/tikz/arccos2.tikz.tex b/buch/papers/ellfilter/tikz/arccos2.tikz.tex
index 3fc3cc6..2cec75f 100644
--- a/buch/papers/ellfilter/tikz/arccos2.tikz.tex
+++ b/buch/papers/ellfilter/tikz/arccos2.tikz.tex
@@ -2,34 +2,45 @@
 
     \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm]
     \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}}
+    \tikzstyle{dot} = [fill, circle, inner sep =0, minimum height=0.1cm]
 
-    \begin{scope}[xscale=0.5]
 
-        \draw[gray, ->] (0,-2) -- (0,2) node[anchor=south]{$\mathrm{Im}~z_1$};
-        \draw[gray, ->] (-10,0) -- (10,0) node[anchor=west]{$\mathrm{Re}~z_1$};
+    \begin{scope}[xscale=0.75]
+
+        \draw[gray, ->] (0,-1) -- (0,2) node[anchor=south]{$\mathrm{Im}~z_1$};
+        \draw[gray, ->] (-2,0) -- (9,0) node[anchor=west]{$\mathrm{Re}~z_1$};
 
         \begin{scope}
 
-            \draw[>->, line width=0.05, thick, blue]   (2, 1.5) -- (2,0.05)  -- node[anchor=south, pos=0.5]{$N=1$} (0.1,0.05) -- (0.1,1.5);
-            \draw[>->, line width=0.05, thick, orange] (4, 1.5) -- (4,0)     -- node[anchor=south, pos=0.25]{$N=2$} (0,0) -- (0,1.5);
-            \draw[>->, line width=0.05, thick, red]    (6, 1.5) node[anchor=north west]{$-\infty$} -- (6,-0.05) node[anchor=west]{$-1$} -- node[anchor=north]{$0$} node[anchor=south, pos=0.1666]{$N=3$} (-0.1,-0.05) node[anchor=east]{$1$}  -- (-0.1,1.5) node[anchor=north east]{$\infty$};
+            \draw[->, ultra thick, blue] (8, 1.5) --  node[align=center]{Sperrbereich} (8,0);
+            \draw[->, ultra thick, cyan] (8, 0) -- node[yshift=-0.5cm]{Durchlassbereich}(4,0);
+            \draw[->, ultra thick, darkgreen] (4, 0) -- node[yshift=-0.5cm]{Durchlassbereich} (0,0);
+            \draw[->, ultra thick, orange] (0, 0) --  node[align=center]{Sperrbereich} (0,1.5);
+
+            \node[anchor=north east] at (8, 1.5) {$-\infty$};
+            \draw (8, 0) node[dot]{} node[anchor=south east] {$1$};
+            \draw (6, 0) node[dot]{} node[anchor=south] {$-1$};
+            \draw (4, 0) node[dot]{} node[anchor=south] {$1$};
+            \draw (2, 0) node[dot]{} node[anchor=south] {$-1$};
+            \draw (0, 0) node[dot]{} node[anchor=south west] {$1$};
+            \node[anchor=north west] at (0, 1.5){$\infty$};
 
+            \node at(4,1) {$N = 4$};
 
-            \node[zero] at (-7,0) {};
-            \node[zero] at (-5,0) {};
-            \node[zero] at (-3,0) {};
+            % \node[zero] at (-7,0) {};
+            % \node[zero] at (-5,0) {};
+            % \node[zero] at (-3,0) {};
             \node[zero] at (-1,0) {};
             \node[zero] at (1,0) {};
             \node[zero] at (3,0) {};
             \node[zero] at (5,0) {};
             \node[zero] at (7,0) {};
 
-
         \end{scope}
 
-        \node[gray, anchor=north] at (-8,0) {$-4\pi$};
-        \node[gray, anchor=north] at (-6,0) {$-3\pi$};
-        \node[gray, anchor=north] at (-4,0) {$-2\pi$};
+        % \node[gray, anchor=north] at (-8,0) {$-4\pi$};
+        % \node[gray, anchor=north] at (-6,0) {$-3\pi$};
+        % \node[gray, anchor=north] at (-4,0) {$-2\pi$};
         \node[gray, anchor=north] at (-2,0) {$-\pi$};
         \node[gray, anchor=north] at (2,0) {$\pi$};
         \node[gray, anchor=north] at (4,0) {$2\pi$};
@@ -37,9 +48,29 @@
         \node[gray, anchor=north] at (8,0) {$4\pi$};
 
 
-        \node[gray, anchor=east] at (0,-1.5) {$-\infty$};
+        % \node[gray, anchor=east] at (0,-1.5) {$-\infty$};
         \node[gray, anchor=east] at (0, 1.5) {$\infty$};
 
     \end{scope}
 
+    \node[zero] at (6.5,2) (n) {};
+    \node[anchor=west] at (n.east) {Nullstelle};
+
+    \begin{scope}[xshift=2.75cm, yshift=-2cm]
+
+        \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$w$};
+
+        \draw[ultra thick, ->, blue]      (-4, 0) -- (-2, 0);
+        \draw[ultra thick, ->, cyan]       (-2, 0) -- (0, 0);
+        \draw[ultra thick, ->, darkgreen]    (0, 0) -- (2, 0);
+        \draw[ultra thick, ->, orange] (2, 0) -- (4, 0);
+
+        \node[anchor=south] at (-4,0) {$-\infty$};
+        \node[anchor=south] at (-2,0) {$-1$};
+        \node[anchor=south] at (0,0) {$0$};
+        \node[anchor=south] at (2,0) {$1$};
+        \node[anchor=south] at (4,0) {$\infty$};
+
+    \end{scope}
+
 \end{tikzpicture}
\ No newline at end of file
diff --git a/buch/papers/ellfilter/tikz/cd.tikz.tex b/buch/papers/ellfilter/tikz/cd.tikz.tex
index 7155a85..0cf2417 100644
--- a/buch/papers/ellfilter/tikz/cd.tikz.tex
+++ b/buch/papers/ellfilter/tikz/cd.tikz.tex
@@ -4,7 +4,7 @@
 
     \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}}
 
-    \begin{scope}[xscale=1, yscale=2]
+    \begin{scope}[xscale=0.9, yscale=1.8]
 
         \draw[gray, ->] (0,-1.5) -- (0,1.5) node[anchor=south]{$\mathrm{Im}~z$};
         \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$\mathrm{Re}~z$};
@@ -22,32 +22,35 @@
 
                 \fill[yellow!30] (0,0) rectangle (1, 0.5);
 
+                \foreach \i in {-2,...,1} {
+                    \foreach \j in {-2,...,1} {
+                        \begin{scope}[xshift=\i*4cm, yshift=\j*1cm]
+                            \draw[->, orange!50] (0, 0) -- (0,0.5);
+                            \draw[->, darkgreen!50] (1, 0) -- (0,0);
+                            \draw[->, cyan!50] (2, 0) -- (1,0);
+                            \draw[->, blue!50] (2,0.5) -- (2, 0);
+                            \draw[->, purple!50] (1, 0.5) -- (2,0.5);
+                            \draw[->, red!50] (0, 0.5) -- (1,0.5);
+                            \draw[->, orange!50] (0,1) -- (0,0.5);
+                            \draw[->, blue!50] (2,0.5) -- (2, 1);
+                            \draw[->, purple!50] (3, 0.5) -- (2,0.5);
+                            \draw[->, red!50] (4, 0.5) -- (3,0.5);
+                            \draw[->, cyan!50] (2, 0) -- (3,0);
+                            \draw[->, darkgreen!50] (3, 0) -- (4,0);
+                        \end{scope}
+                    }
+                }
 
-                \draw[thick, ->, darkgreen] (0, 0) -- (0,0.5);
-                \draw[thick, ->, orange] (1, 0) -- (0,0);
-                \draw[thick, ->, red] (2, 0) -- (1,0);
-                \draw[thick, ->, blue] (2,0.5) -- (2, 0);
-                \draw[thick, ->, purple] (1, 0.5) -- (2,0.5);
-                \draw[thick, ->, cyan] (0, 0.5) -- (1,0.5);
-
-
+                \draw[ultra thick, ->, orange] (0, 0) -- (0,0.5);
+                \draw[ultra thick, ->, darkgreen] (1, 0) -- (0,0);
+                \draw[ultra thick, ->, cyan] (2, 0) -- (1,0);
+                \draw[ultra thick, ->, blue] (2,0.5) -- (2, 0);
+                \draw[ultra thick, ->, purple] (1, 0.5) -- (2,0.5);
+                \draw[ultra thick, ->, red] (0, 0.5) -- (1,0.5);
 
                 \foreach \i in {-2,...,1} {
                     \foreach \j in {-2,...,1} {
                         \begin{scope}[xshift=\i*4cm, yshift=\j*1cm]
-                            \draw[opacity=0.5, ->, darkgreen] (0, 0) -- (0,0.5);
-                            \draw[opacity=0.5, ->, orange] (1, 0) -- (0,0);
-                            \draw[opacity=0.5, ->, red] (2, 0) -- (1,0);
-                            \draw[opacity=0.5, ->, blue] (2,0.5) -- (2, 0);
-                            \draw[opacity=0.5, ->, purple] (1, 0.5) -- (2,0.5);
-                            \draw[opacity=0.5, ->, cyan] (0, 0.5) -- (1,0.5);
-                            \draw[opacity=0.5, ->, darkgreen] (0,1) -- (0,0.5);
-                            \draw[opacity=0.5, ->, blue] (2,0.5) -- (2, 1);
-                            \draw[opacity=0.5, ->, purple] (3, 0.5) -- (2,0.5);
-                            \draw[opacity=0.5, ->, cyan] (4, 0.5) -- (3,0.5);
-                            \draw[opacity=0.5, ->, red] (2, 0) -- (3,0);
-                            \draw[opacity=0.5, ->, orange] (3, 0) -- (4,0);
-
                             \node[zero] at ( 1, 0) {};
                             \node[zero] at ( 3, 0) {};
                             \node[pole] at ( 1,0.5) {};
@@ -63,16 +66,21 @@
 
     \end{scope}
 
-    \begin{scope}[yshift=-3.5cm, xscale=0.75]
+    \node[zero] at (4,3) (n) {};
+    \node[anchor=west] at (n.east) {Nullstelle};
+    \node[pole, below=0.25cm of n] (n) {};
+    \node[anchor=west] at (n.east) {Polstelle};
+
+    \begin{scope}[yshift=-4cm, xscale=0.75]
 
         \draw[gray, ->] (-6,0) -- (6,0) node[anchor=west]{$w$};
 
-        \draw[thick, ->, purple] (-5, 0) -- (-3, 0);
-        \draw[thick, ->, blue]      (-3, 0) -- (-2, 0);
-        \draw[thick, ->, red]       (-2, 0) -- (0, 0);
-        \draw[thick, ->, orange]    (0, 0) -- (2, 0);
-        \draw[thick, ->, darkgreen] (2, 0) -- (3, 0);
-        \draw[thick, ->, cyan] (3, 0) -- (5, 0);
+        \draw[ultra thick, ->, purple] (-5, 0) -- (-3, 0);
+        \draw[ultra thick, ->, blue]      (-3, 0) -- (-2, 0);
+        \draw[ultra thick, ->, cyan]       (-2, 0) -- (0, 0);
+        \draw[ultra thick, ->, darkgreen]    (0, 0) -- (2, 0);
+        \draw[ultra thick, ->, orange] (2, 0) -- (3, 0);
+        \draw[ultra thick, ->, red] (3, 0) -- (5, 0);
 
         \node[anchor=south] at (-5,0) {$-\infty$};
         \node[anchor=south] at (-3,0) {$-1/k$};
diff --git a/buch/papers/ellfilter/tikz/cd2.tikz.tex b/buch/papers/ellfilter/tikz/cd2.tikz.tex
index 0743f7d..d4187c4 100644
--- a/buch/papers/ellfilter/tikz/cd2.tikz.tex
+++ b/buch/papers/ellfilter/tikz/cd2.tikz.tex
@@ -5,9 +5,9 @@
 
     \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}}
 
-    \begin{scope}[xscale=1.25, yscale=2.5]
+    \begin{scope}[xscale=1.25, yscale=3.5]
 
-        \draw[gray, ->] (0,-0.75) -- (0,1.25) node[anchor=south]{$\mathrm{Im}~z_1$};
+        \draw[gray, ->] (0,-0.55) -- (0,1.05) node[anchor=south]{$\mathrm{Im}~z_1$};
         \draw[gray, ->] (-1.5,0) -- (6,0) node[anchor=west]{$\mathrm{Re}~z_1$};
 
         \draw[gray] ( 1,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K_1$};
@@ -35,18 +35,30 @@
             % \node[] at (2.5, 0.25) {\small $N=3$};
 
             \fill[orange!30] (0,0) rectangle (5, 0.5);
-            \fill[yellow!30] (0,0) rectangle (1, 0.5);
+            % \fill[yellow!30] (0,0) rectangle (1, 0.1);
             \node[] at (2.5, 0.25) {\small $N=5$};
 
 
             \draw[decorate,decoration={brace,amplitude=3pt,mirror}, yshift=0.05cm]
-                (5,0.5) node(t_k_unten){} -- node[above, yshift=0.1cm]{$NK$}
+                (5,0.5) node(t_k_unten){} -- node[above, yshift=0.1cm]{$NK_1$}
                 (0,0.5) node(t_k_opt_unten){};
 
             \draw[decorate,decoration={brace,amplitude=3pt,mirror}, xshift=0.1cm]
                 (5,0) node(t_k_unten){} -- node[right, xshift=0.1cm]{$K^\prime \frac{K_1N}{K} = K^\prime_1$}
                 (5,0.5) node(t_k_opt_unten){};
 
+
+            \draw[ultra thick, ->, darkgreen] (5, 0) -- node[yshift=-0.5cm]{Durchlassbereich} (0,0);
+            \draw[ultra thick, ->, orange] (-0, 0) --  node[align=center]{Übergangs-\\berech} (0,0.5);
+            \draw[ultra thick, ->, red] (0,0.5) -- node[align=center, yshift=0.7cm]{Sperrbereich} (5, 0.5);
+
+            \draw (4,0  )  node[dot]{} node[anchor=south]      {\small $1$};
+            \draw (2,0  )  node[dot]{} node[anchor=south]      {\small $-1$};
+            \draw (0,0  )  node[dot]{} node[anchor=south west] {\small $1$};
+            \draw (0,0.5)  node[dot]{} node[anchor=north west] {\small $1/k$};
+            \draw (2,0.5)  node[dot]{} node[anchor=north]      {\small $-1/k$};
+            \draw (4,0.5)  node[dot]{} node[anchor=north]      {\small $1/k$};
+
             \foreach \i in {-2,...,1} {
                 \foreach \j in {-2,...,1} {
                     \begin{scope}[xshift=\i*4cm, yshift=\j*1cm]
@@ -60,24 +72,22 @@
                 }
             }
 
+        \end{scope}
 
+    \end{scope}
 
+    \begin{scope}[xshift=1cm , yshift=-3cm, xscale=0.75]
 
-            \draw[thick, ->, darkgreen] (5, 0) -- node[yshift=-0.5cm]{Durchlassbereich} (0,0);
-            \draw[thick, ->, orange] (-0, 0) --  node[align=center]{Übergangs-\\berech} (0,0.5);
-            \draw[thick, ->, red] (0,0.5) -- node[align=center, yshift=0.5cm]{Sperrbereich} (5, 0.5);
-
-            \draw (4,0  )  node[dot]{} node[anchor=south]      {\small $1$};
-            \draw (2,0  )  node[dot]{} node[anchor=south]      {\small $-1$};
-            \draw (0,0  )  node[dot]{} node[anchor=south west] {\small $1$};
-            \draw (0,0.5)  node[dot]{} node[anchor=north west] {\small $1/k$};
-            \draw (2,0.5)  node[dot]{} node[anchor=north]      {\small $-1/k$};
-            \draw (4,0.5)  node[dot]{} node[anchor=north]      {\small $1/k$};
-
-
+        \draw[gray, ->] (-1,0) -- (6,0) node[anchor=west]{$w$};
 
-        \end{scope}
+        \draw[ultra thick, ->, darkgreen]    (0, 0) -- (2, 0);
+        \draw[ultra thick, ->, orange] (2, 0) -- (3, 0);
+        \draw[ultra thick, ->, red] (3, 0) -- (5, 0);
 
+        \node[anchor=south] at (0,0) {$0$};
+        \node[anchor=south] at (2,0) {$1$};
+        \node[anchor=south] at (3,0) {$1/k$};
+        \node[anchor=south] at (5,0) {$\infty$};
 
     \end{scope}
 
diff --git a/buch/papers/ellfilter/tikz/cd3.tikz.tex b/buch/papers/ellfilter/tikz/cd3.tikz.tex
new file mode 100644
index 0000000..ae18519
--- /dev/null
+++ b/buch/papers/ellfilter/tikz/cd3.tikz.tex
@@ -0,0 +1,86 @@
+\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2]
+
+    \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm]
+    \tikzstyle{dot} = [fill, circle, inner sep =0, minimum height=0.1cm]
+
+    \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}}
+
+    \begin{scope}[xscale=1.25, yscale=2.5]
+
+        \draw[gray, ->] (0,-0.55) -- (0,1.05) node[anchor=south]{$\mathrm{Im}$};
+        \draw[gray, ->] (-1.5,0) -- (6,0) node[anchor=west]{$\mathrm{Re}$};
+
+        % \draw[gray] ( 1,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K_1$};
+        % \draw[gray] ( 5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $5K_1$};
+        % \draw[gray]  (0, 0.5) +(0.1, 0) -- +(-0.1, 0) node[inner sep=0, anchor=east]{\small $jK^\prime_1$};
+
+        \begin{scope}
+
+            \clip(-1.5,-0.75) rectangle (6.8,1.25);
+
+            % \draw[>->, line width=0.05, thick, blue]   (1, 0.45) -- (2, 0.45) -- (2, 0.05) -- ( 0.1, 0.05) -- ( 0.1,0.45) -- (1, 0.45);
+            % \draw[>->, line width=0.05, thick, orange] (2, 0.5 ) -- (4, 0.5 ) -- (4, 0   ) -- ( 0  , 0   ) -- ( 0  ,0.5 ) -- (2, 0.5 );
+            % \draw[>->, line width=0.05, thick, red]    (3, 0.55) -- (6, 0.55) -- (6,-0.05) -- (-0.1,-0.05) -- (-0.1,0.55) -- (3, 0.55);
+            % \node[blue] at (1, 0.25) {$N=1$};
+            % \node[orange] at (3, 0.25) {$N=2$};
+            % \node[red] at (5, 0.25) {$N=3$};
+
+
+
+            % \draw[line width=0.1cm, fill, red!50] (0,0) rectangle (3, 0.5);
+            % \draw[line width=0.05cm, fill, orange!50] (0,0) rectangle (2, 0.5);
+            % \fill[yellow!50] (0,0) rectangle (1, 0.5);
+            % \node[] at (0.5, 0.25) {\small $N=1$};
+            % \node[] at (1.5, 0.25) {\small $N=2$};
+            % \node[] at (2.5, 0.25) {\small $N=3$};
+
+            % \fill[orange!30] (0,0) rectangle (5, 0.5);
+            \fill[yellow!30] (0,0) rectangle (1, 0.5);
+
+
+            % \draw[decorate,decoration={brace,amplitude=3pt,mirror}, yshift=0.05cm]
+            %     (5,0.5) node(t_k_unten){} -- node[above, yshift=0.1cm]{$NK_1$}
+            %     (0,0.5) node(t_k_opt_unten){};
+
+            % \draw[decorate,decoration={brace,amplitude=3pt,mirror}, xshift=0.1cm]
+            %     (5,0) node(t_k_unten){} -- node[right, xshift=0.1cm]{$K^\prime \frac{K_1N}{K} = K^\prime_1$}
+            %     (5,0.5) node(t_k_opt_unten){};
+
+            \foreach \i in {-2,...,1} {
+                \foreach \j in {-2,...,1} {
+                    \begin{scope}[xshift=\i*4cm, yshift=\j*1cm]
+
+                        \node[zero] at ( 1, 0) {};
+                        \node[zero] at ( 3, 0) {};
+                        \node[pole] at ( 1,0.5) {};
+                        \node[pole] at ( 3,0.5) {};
+
+                    \end{scope}
+                }
+            }
+
+
+
+            \onslide<2->{
+            \draw[ultra thick, ->, darkgreen] (5, 0) -- node[yshift=-0.4cm]{Durchlassbereich} (0,0);
+            \draw[ultra thick, ->, orange] (-0, 0) --  node[align=center]{Übergangs-\\berech} (0,0.5);
+            \draw[ultra thick, ->, red] (0,0.5) -- node[align=center, yshift=0.4cm]{Sperrbereich} (5, 0.5);
+            \node[] at (2.5, 0.25) {\small $N=5$};
+            }
+            \onslide<1->{
+            \draw (4,0  )  node[dot]{} node[anchor=south]      {\small $1$};
+            \draw (2,0  )  node[dot]{} node[anchor=south]      {\small $-1$};
+            \draw (0,0  )  node[dot]{} node[anchor=south west] {\small $1$};
+            \draw (0,0.5)  node[dot]{} node[anchor=north west] {\small $1/k$};
+            \draw (2,0.5)  node[dot]{} node[anchor=north]      {\small $-1/k$};
+            \draw (4,0.5)  node[dot]{} node[anchor=north]      {\small $1/k$};
+
+            }
+
+
+        \end{scope}
+
+
+    \end{scope}
+
+\end{tikzpicture}
\ No newline at end of file
diff --git a/buch/papers/ellfilter/tikz/elliptic_transform1.tikz.tex b/buch/papers/ellfilter/tikz/elliptic_transform1.tikz.tex
new file mode 100644
index 0000000..2a36ee0
--- /dev/null
+++ b/buch/papers/ellfilter/tikz/elliptic_transform1.tikz.tex
@@ -0,0 +1,76 @@
+\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2]
+
+    \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm]
+
+    \tikzset{pole/.style={cross out, draw, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}}
+
+    \begin{scope}[xscale=1, yscale=1.5]
+
+    \begin{scope}[]
+
+        \fill[orange!25] (0,0) rectangle (1.5, 0.75);
+        \fill[yellow!50] (0,0) rectangle (0.5, 0.25);
+
+        \draw[gray, ->] (0,-0.75) -- (0,1.25) node[anchor=south]{$\mathrm{Im}~z$};
+        \draw[gray, ->] (-1.75,0) -- (1.75,0) node[anchor=west]{$\mathrm{Re}~z$};
+
+        \draw[gray] ( 0.5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K$};
+        \draw[gray]  (0, 0.25) +(0.05, 0) -- +(-0.05, 0) node[inner sep=0, anchor=east]{\small $jK^\prime$};
+
+        % \draw[gray] ( 1.5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K_1$};
+        % \draw[gray]  (0, 0.75) +(0.05, 0) -- +(-0.05, 0) node[inner sep=0, anchor=east]{\small $jK_1^\prime$};
+
+        \clip(-1.6,-0.6) rectangle (1.6,1.6);
+        \begin{scope}[xscale=0.5, yscale=0.25, blue]
+            \foreach \i in {-1,...,1} {
+                \foreach \j in {-1,...,2} {
+                    \begin{scope}[xshift=\i*2cm, yshift=\j*2cm]
+                        \node[zero] at ( 1, 0) {};
+                        \node[zero] at ( -1, 0) {};
+                        \node[pole] at ( 1,1) {};
+                        \node[pole] at ( -1,1) {};
+                    \end{scope}
+                }
+            }
+        \end{scope}
+
+        \node at (0,2) {$\cd \left(N~K_1~z , k_1 \right)$};
+        \node at (0,2) {$w= \cd(z K, k)$};
+
+        \draw[scale=0.2, domain=0.02:5,  variable=\x, red] plot ({\x1+3}, {1/\x+2});
+
+    \end{scope}
+
+    \begin{scope}[xshift=5cm]
+
+        \fill[orange!50] (0,0) rectangle (1.5, 0.75);
+        \fill[yellow!25] (0,0) rectangle (0.5, 0.25);
+
+        \draw[gray, ->] (0,-0.75) -- (0,1.25) node[anchor=south]{$\mathrm{Im}~z$};
+        \draw[gray, ->] (-1.75,0) -- (1.75,0) node[anchor=west]{$\mathrm{Re}~z$};
+
+        % \draw[gray] ( 0.5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K$};
+        % \draw[gray]  (0, 0.25) +(0.05, 0) -- +(-0.05, 0) node[inner sep=0, anchor=east]{\small $jK^\prime$};
+
+        \draw[gray] ( 0.5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K_1$};
+        \draw[gray]  (0, 0.75) +(0.05, 0) -- +(-0.05, 0) node[inner sep=0, anchor=east]{\small $jK_1^\prime$};
+
+        \clip(-1.6,-0.6) rectangle (1.6,1.6);
+        \begin{scope}[xscale=0.5, yscale=0.75, red]
+            \foreach \i in {-1,...,1} {
+                \foreach \j in {-1,...,0} {
+                    \begin{scope}[xshift=\i*2cm, yshift=\j*2cm]
+                        \node[zero] at ( 1, 0) {};
+                        \node[zero] at ( -1, 0) {};
+                        \node[pole] at ( 1,1) {};
+                        \node[pole] at ( -1,1) {};
+                    \end{scope}
+                }
+            }
+        \end{scope}
+
+    \end{scope}
+
+\end{scope}
+
+\end{tikzpicture}
diff --git a/buch/papers/ellfilter/tikz/elliptic_transform2.tikz.tex b/buch/papers/ellfilter/tikz/elliptic_transform2.tikz.tex
new file mode 100644
index 0000000..20c2d82
--- /dev/null
+++ b/buch/papers/ellfilter/tikz/elliptic_transform2.tikz.tex
@@ -0,0 +1,75 @@
+
+\def\d{0.2}
+\def\n{3}
+\def\nn{2}
+\def\a{2.5}
+
+\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2]
+
+    \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm]
+    \tikzstyle{dot} = [fill, circle, inner sep =0, minimum height=0.1cm]
+
+    \tikzset{pole/.style={cross out, draw, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}}
+
+    \begin{scope}[xscale=3, yscale=3]
+
+    \begin{scope}[]
+        % \onslide<4->{
+        \fill[orange!30, scale=1.735] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5);
+        % }
+        % \onslide<2->{
+        \fill[yellow!30] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5);
+        % }
+
+        \begin{scope}[]
+            \clip(0,0) rectangle (2,1.25);
+            \draw[thick, scale=1, domain=0.1:10,  variable=\x, smooth, samples=200] plot ({\d*\x1+0.5}, {\d/\x+0.5});
+            \node at(1.25,0.7) {$K + jK^\prime$ Ortskurve};
+        \end{scope}
+
+        % \onslide<2->{
+        \begin{scope}[blue]
+            \draw[] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5);
+
+
+            \node[pole] at ( \d*\a+0.5, \d/\a+0.5) {};
+            \node[zero] at ( \d*\a+0.5, 0) {};
+
+            \draw[] ( \d*\a+0.5,0)  node[anchor=north] {\small $K$};
+            \draw[]  (0, \d/\a+0.5)  node[anchor=east]{\small $jK^\prime$};
+
+            % \onslide<3->{
+
+            \foreach \i in {1,...,\nn} {
+                \draw[gray, dotted] (\i*\d*\a/\n+\i*0.5/\n, 0) -- (\i*\d*\a/\n+\i*0.5/\n, \d/\a+0.5);
+            }
+
+            \node[dot, gray] at (\d*\a/\n+0.5/\n, \d/\a+0.5) {};
+            \node[above] at (0.5*\d*\a/\n+0.5*0.5/\n, \d/\a+0.5) {\small $K/N$};
+            % }
+        \end{scope}
+        % }
+
+        % \onslide<4->{
+        \begin{scope}[scale=1.735, red]
+            \draw (0,0) rectangle (\d*\a/\n+0.5/\n, \d/\a+0.5);
+            \draw[gray] (0,0) -- (\d*\a/\n+0.5/\n, \d/\a+0.5);
+
+            \node[pole] at ( \d*\a/\n+0.5/\n, \d/\a+0.5) {};
+            \node[zero] at ( \d*\a/\n+0.5/\n, 0) {};
+
+
+            \draw[] ( \d*\a/\n+0.5/\n,0)  node[anchor=north] {\small $K_1$};
+            \draw[]  (0, \d/\a+0.5)  node[anchor=east]{\small $jK_1^\prime$};
+
+        \end{scope}
+        % }
+
+        \draw[gray, ->] (0,-0.25) -- (0,1.25) node[anchor=south]{$\mathrm{Im}$};
+        \draw[gray, ->] (-0.25,0) -- (2,0) node[anchor=west]{$\mathrm{Re}$};
+
+    \end{scope}
+
+\end{scope}
+
+\end{tikzpicture}
diff --git a/buch/papers/ellfilter/tikz/filter.tikz.tex b/buch/papers/ellfilter/tikz/filter.tikz.tex
new file mode 100644
index 0000000..769602a
--- /dev/null
+++ b/buch/papers/ellfilter/tikz/filter.tikz.tex
@@ -0,0 +1,32 @@
+\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2]
+
+    \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm]
+
+    \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}}
+
+    \begin{scope}[xscale=3, yscale=2.5]
+
+        \fill[darkgreen!15] (0,0) rectangle  (1,1);
+        \node[darkgreen] at (0.5,0.5) {Durchlassbereich};
+        \fill[orange!15] (1,0) rectangle  (2.5,1);
+        \node[orange] at (1.75,0.5) {Sperrbereich};
+
+        \draw[gray, ->] (0,0) -- (0,1.25) node[anchor=south]{$|H(\Omega)|$};
+        \draw[gray, ->] (0,0) -- (2.75,0) node[anchor=west]{$\Omega$};
+
+        \draw[dashed] (0,0.707) node[left] {$\sqrt{\frac{1}{1+\varepsilon^2}}$} -| (1,0) node[below] {$\Omega_p$};
+        \draw[dashed] (0,0.707) node[left] {$\sqrt{\frac{1}{1+\varepsilon^2}}$} -| (1,0) node[below] {$\Omega_p$};
+
+        \node[left] at(0,1) {$1$};
+
+        \draw[red, thick] (0,1) -- (1,1) -- (1,0) -- (2.5,0);
+
+        \node[anchor=north, red] at (0.5,1) {Ideal};
+
+        \draw[thick, domain=0:2.5,  variable=\x, smooth, samples=200] plot
+        ({\x}, {sqrt(abs(1/ (1 + \x^10)))});
+        \node[anchor=south] at (0.5,1) {Butterworth ($N=5$)};
+
+    \end{scope}
+
+\end{tikzpicture}
diff --git a/buch/papers/ellfilter/tikz/sn.tikz.tex b/buch/papers/ellfilter/tikz/sn.tikz.tex
index 87c63c0..0546fda 100644
--- a/buch/papers/ellfilter/tikz/sn.tikz.tex
+++ b/buch/papers/ellfilter/tikz/sn.tikz.tex
@@ -4,7 +4,7 @@
 
     \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}}
 
-    \begin{scope}[xscale=1, yscale=2]
+    \begin{scope}[xscale=0.9, yscale=1.8]
 
         \draw[gray, ->] (0,-1.5) -- (0,1.5) node[anchor=south]{$\mathrm{Im}~z$};
         \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$\mathrm{Re}~z$};
@@ -17,35 +17,45 @@
 
             \begin{scope}[xshift=-1cm]
 
-                \draw[thick, ->, darkgreen] (0, 0) -- (0,0.5);
-                \draw[thick, ->, orange] (1, 0) -- (0,0);
-                \draw[thick, ->, red] (2, 0) -- (1,0);
-                \draw[thick, ->, blue] (2,0.5) -- (2, 0);
-                \draw[thick, ->, purple] (1, 0.5) -- (2,0.5);
-                \draw[thick, ->, cyan] (0, 0.5) -- (1,0.5);
+                \foreach \i in {-2,...,2} {
+                    \foreach \j in {-2,...,1} {
+                        \begin{scope}[xshift=\i*4cm, yshift=\j*1cm]
+                            \draw[<-, blue!50] (0, 0) -- (0,0.5);
+                            \draw[<-, cyan!50] (1, 0) -- (0,0);
+                            \draw[<-, darkgreen!50] (2, 0) -- (1,0);
+                            \draw[<-, orange!50] (2,0.5) -- (2, 0);
+                            \draw[<-, red!50] (1, 0.5) -- (2,0.5);
+                            \draw[<-, purple!50] (0, 0.5) -- (1,0.5);
+                            \draw[<-, blue!50] (0,1) -- (0,0.5);
+                            \draw[<-, orange!50] (2,0.5) -- (2, 1);
+                            \draw[<-, red!50] (3, 0.5) -- (2,0.5);
+                            \draw[<-, purple!50] (4, 0.5) -- (3,0.5);
+                            \draw[<-, darkgreen!50] (2, 0) -- (3,0);
+                            \draw[<-, cyan!50] (3, 0) -- (4,0);
+                        \end{scope}
+                    }
+                }
+
+                % \pause
+                \draw[ultra thick, <-, darkgreen] (2, 0) -- (1,0);
+                % \pause
+                \draw[ultra thick, <-, orange] (2,0.5) -- (2, 0);
+                % \pause
+                \draw[ultra thick, <-, red] (1, 0.5) -- (2,0.5);
+                % \pause
+                \draw[ultra thick, <-, blue] (0, 0) -- (0,0.5);
+                \draw[ultra thick, <-, purple] (0, 0.5) -- (1,0.5);
+                \draw[ultra thick, <-, cyan] (1, 0) -- (0,0);
+                % \pause
 
 
                 \foreach \i in {-2,...,2} {
                     \foreach \j in {-2,...,1} {
                         \begin{scope}[xshift=\i*4cm, yshift=\j*1cm]
-                            \draw[opacity=0.5, ->, darkgreen] (0, 0) -- (0,0.5);
-                            \draw[opacity=0.5, ->, orange] (1, 0) -- (0,0);
-                            \draw[opacity=0.5, ->, red] (2, 0) -- (1,0);
-                            \draw[opacity=0.5, ->, blue] (2,0.5) -- (2, 0);
-                            \draw[opacity=0.5, ->, purple] (1, 0.5) -- (2,0.5);
-                            \draw[opacity=0.5, ->, cyan] (0, 0.5) -- (1,0.5);
-                            \draw[opacity=0.5, ->, darkgreen] (0,1) -- (0,0.5);
-                            \draw[opacity=0.5, ->, blue] (2,0.5) -- (2, 1);
-                            \draw[opacity=0.5, ->, purple] (3, 0.5) -- (2,0.5);
-                            \draw[opacity=0.5, ->, cyan] (4, 0.5) -- (3,0.5);
-                            \draw[opacity=0.5, ->, red] (2, 0) -- (3,0);
-                            \draw[opacity=0.5, ->, orange] (3, 0) -- (4,0);
-
                             \node[zero] at ( 1, 0) {};
                             \node[zero] at ( 3, 0) {};
                             \node[pole] at ( 1,0.5) {};
                             \node[pole] at ( 3,0.5) {};
-
                         \end{scope}
                     }
                 }
@@ -57,20 +67,23 @@
         \draw[gray] ( 1,0) +(0,0.1) -- +(0, -0.1) node[inner sep=0, anchor=north] {\small $K$};
         \draw[gray]  (0, 0.5) +(0.1, 0) -- +(-0.1, 0) node[inner sep=0, anchor=east]{\small $jK^\prime$};
 
-
-
     \end{scope}
 
-    \begin{scope}[yshift=-3.5cm, xscale=0.75]
+    \node[zero] at (4,3) (n) {};
+    \node[anchor=west] at (n.east) {Nullstelle};
+    \node[pole, below=0.25cm of n] (n) {};
+    \node[anchor=west] at (n.east) {Polstelle};
+
+    \begin{scope}[yshift=-4cm, xscale=0.75]
 
         \draw[gray, ->] (-6,0) -- (6,0) node[anchor=west]{$w$};
 
-        \draw[thick, ->, purple] (-5, 0) -- (-3, 0);
-        \draw[thick, ->, blue]      (-3, 0) -- (-2, 0);
-        \draw[thick, ->, red]       (-2, 0) -- (0, 0);
-        \draw[thick, ->, orange]    (0, 0) -- (2, 0);
-        \draw[thick, ->, darkgreen] (2, 0) -- (3, 0);
-        \draw[thick, ->, cyan] (3, 0) -- (5, 0);
+        \draw[ultra thick, ->, purple] (-5, 0) -- (-3, 0);
+        \draw[ultra thick, ->, blue]      (-3, 0) -- (-2, 0);
+        \draw[ultra thick, ->, cyan]       (-2, 0) -- (0, 0);
+        \draw[ultra thick, ->, darkgreen]    (0, 0) -- (2, 0);
+        \draw[ultra thick, ->, orange] (2, 0) -- (3, 0);
+        \draw[ultra thick, ->, red] (3, 0) -- (5, 0);
 
         \node[anchor=south] at (-5,0) {$-\infty$};
         \node[anchor=south] at (-3,0) {$-1/k$};
diff --git a/buch/papers/ellfilter/tschebyscheff.tex b/buch/papers/ellfilter/tschebyscheff.tex
index 7d426b6..639c87c 100644
--- a/buch/papers/ellfilter/tschebyscheff.tex
+++ b/buch/papers/ellfilter/tschebyscheff.tex
@@ -1,9 +1,8 @@
 \section{Tschebyscheff-Filter}
 
-Als Einstieg betrachent Wir das Tschebyscheff-Filter, welches sehr verwand ist mit dem elliptischen Filter.
-Genauer ausgedrückt sind die Tschebyscheff-1 und -2 Filter Spezialfälle davon.
-
-Der Name des Filters deutet schon an, dass die Tschebyscheff-Polynome $T_N$ für das Filter relevant sind:
+Als Einstieg betrachten wir das Tschebyscheff-Filter, welches sehr verwandt ist mit dem elliptischen Filter.
+Genauer ausgedrückt erhält man die Tschebyscheff-1 und -2 Filter bei Grenzwerten von Parametern beim elliptischen Filter.
+Der Name des Filters deutet schon an, dass die Tschebyscheff-Polynome $T_N$ (siehe auch Kapitel \label{buch:polynome:section:tschebyscheff}) für das Filter relevant sind:
 \begin{align}
     T_{0}(x)&=1\\
     T_{1}(x)&=x\\
@@ -16,7 +15,7 @@ Bemerkenswert ist, dass die Polynome im Intervall $[-1, 1]$ mit der trigonometri
     T_N(w) &= \cos \left( N \cos^{-1}(w) \right) \\
            &= \cos \left(N~z \right), \quad w= \cos(z)
 \end{align}
-übereinstimmt.
+übereinstimmen.
 Der Zusammenhang lässt sich mit den Doppel- und Mehrfachwinkelfunktionen der trigonometrischen Funktionen erklären.
 Abbildung \ref{ellfilter:fig:chebychef_polynomials} zeigt einige Tschebyscheff-Polynome.
 \begin{figure}
@@ -28,7 +27,7 @@ Abbildung \ref{ellfilter:fig:chebychef_polynomials} zeigt einige Tschebyscheff-P
 Da der Kosinus begrenzt zwischen $-1$ und $1$ ist, sind auch die Tschebyscheff-Polynome begrenzt.
 Geht man aber über das Intervall $[-1, 1]$ hinaus, divergieren die Funktionen mit zunehmender Ordnung immer steiler gegen $\pm \infty$.
 Diese Eigenschaft ist sehr nützlich für ein Filter.
-Wenn wir die Tschebyscheff-Polynome quadrieren, passen sie perfekt in die Voraussetzungen für Filterfunktionen, wie es Abbildung \ref{ellfiter:fig:chebychef} demonstriert.
+Wenn wir die Tschebyscheff-Polynome quadrieren, passen sie perfekt in die Forderungen für Filterfunktionen, wie es Abbildung \ref{ellfiter:fig:chebychef} demonstriert.
 \begin{figure}
     \centering
     \input{papers/ellfilter/python/F_N_chebychev.pgf}
@@ -36,12 +35,11 @@ Wenn wir die Tschebyscheff-Polynome quadrieren, passen sie perfekt in die Voraus
     \label{ellfiter:fig:chebychef}
 \end{figure}
 
-
 Die analytische Fortsetzung von \eqref{ellfilter:eq:chebychef_polynomials} über das Intervall $[-1,1]$ hinaus stimmt mit den Polynomen überein, wie es zu erwarten ist.
-Die genauere Betrachtung wird uns dann helfen die elliptischen Filter besser zu verstehen.
+Die genauere Betrachtung wird uns helfen die elliptischen Filter besser zu verstehen.
 
-Starten wir mit der Funktion, die als erstes auf $w$ angewendet wird, dem Arcuscosinus.
-Die invertierte Funktion des Kosinus kann als definites Integral dargestellt werden:
+Starten wir mit der Funktion, die in \eqref{ellfilter:eq:chebychef_polynomials} als erstes auf $w$ angewendet wird, dem Arcuscosinus.
+Die invertierte Funktion des Kosinus kann als bestimmtes Integral dargestellt werden:
 \begin{align}
     \cos^{-1}(x)
     &=
@@ -63,9 +61,9 @@ Die invertierte Funktion des Kosinus kann als definites Integral dargestellt wer
         }
     }
     ~dz
-    + \frac{\pi}{2}
+    + \frac{\pi}{2}.
 \end{align}
-Der Integrand oder auch die Ableitung
+Der Integrand oder auch die Ableitung von $\cos^{-1}(x)$
 \begin{equation}
     \frac{
         -1
@@ -75,59 +73,37 @@ Der Integrand oder auch die Ableitung
         }
     }
 \end{equation}
-bestimmt dabei die Richtung, in der die Funktion verläuft.
+bestimmt dabei die Richtung, in welche die Funktion verläuft.
 Der reelle Arcuscosinus is bekanntlich nur für $|z| \leq 1$ definiert.
 Hier bleibt der Wert unter der Wurzel positiv und das Integral liefert reelle Werte.
 Doch wenn $|z|$ über 1 hinausgeht, wird der Term unter der Wurzel negativ.
 Durch die Quadratwurzel entstehen für den Integranden zwei rein komplexe Lösungen.
 Der Wert des Arcuscosinus verlässt also bei $z= \pm 1$ den reellen Zahlenstrahl und knickt in die komplexe Ebene ab.
-Abbildung \ref{ellfilter:fig:arccos} zeigt den $\arccos$ in der komplexen Ebene.
+Abbildung \ref{ellfilter:fig:arccos} zeigt den Arcuscosinus in der komplexen Ebene.
 \begin{figure}
     \centering
     \input{papers/ellfilter/tikz/arccos.tikz.tex}
     \caption{Die Funktion $z = \cos^{-1}(w)$ dargestellt in der komplexen ebene.}
     \label{ellfilter:fig:arccos}
 \end{figure}
-Wegen der Periodizität des Kosinus ist auch der Arcuscosinus $2\pi$-periodisch und es entstehen periodische Nullstellen.
-% \begin{equation}
-%     \frac{
-%         1
-%     }{
-%         \sqrt{
-%             1-z^2
-%         }
-%     }
-%     \in \mathbb{R}
-%     \quad
-%     \forall
-%     \quad
-%     -1  \leq z \leq 1
-% \end{equation}
-% \begin{equation}
-%     \frac{
-%         1
-%     }{
-%         \sqrt{
-%             1-z^2
-%         }
-%     }
-%     = i \xi \quad | \quad \xi \in \mathbb{R}
-%     \quad
-%     \forall
-%     \quad
-%     z \leq -1 \cup z \geq 1
-% \end{equation}
+Wegen der Periodizität des Kosinus ist auch der Arcuscosinus $2\pi$-periodisch.
+Das Einzeichnen von Pol- und Nullstellen ist hilfreich für die Betrachtung der Funktion.
+
 
-Die Tschebyscheff-Polynome skalieren diese Nullstellen mit dem Ordnungsfaktor $N$, wie dargestellt in Abbildung \ref{ellfilter:fig:arccos2}.
+In \eqref{ellfilter:eq:chebychef_polynomials} wird $z$ mit dem Ordnungsfaktor $N$ multipliziert und durch die Kosinusfunktion zurück transformiert.
+Die Skalierung hat zur folge, dass bei der Rücktransformation durch den Kosinus mehrere Nullstellen durchlaufen werden.
+Somit passiert $\cos( N~\cos^{-1}(w))$ im Intervall $[-1, 1]$ $N$ Nullstellen, wie dargestellt in Abbildung \ref{ellfilter:fig:arccos2}.
 \begin{figure}
     \centering
     \input{papers/ellfilter/tikz/arccos2.tikz.tex}
     \caption{
         $z_1=N \cos^{-1}(w)$-Ebene der Tschebyscheff-Funktion.
-        Die eingefärbten Pfade sind Verläufe von $w~\forall~[-\infty, \infty]$ für verschiedene Ordnungen $N$.
-        Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen werden passiert.
+        Die eingefärbten Pfade sind Verläufe von $w\in(-\infty, \infty)$ für $N = 4$.
+        Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen werden passiert die zu Equirippel-Verhalten führen.
+        Die vertikalen Segmente der Funktion sorgen für das Ansteigen der Funktion gegen $\infty$ nach der Grenzfrequenz.
+        Die eingezeichneten Nullstellen sind vom zurücktransformierenden Kosinus.
     }
     \label{ellfilter:fig:arccos2}
 \end{figure}
-Somit passert $\cos( N~\cos^{-1}(w))$ im Intervall $[-1, 1]$ $N$ Nullstellen.
-Durch die spezielle Anordnung der Nullstellen hat die Funktion Equirippel-Verhalten und ist dennoch ein Polynom, was sich perfekt für linear Filter eignet.
+Durch die spezielle Anordnung der Nullstellen hat die Funktion auf der reellen Achse Equirippel-Verhalten und ist dennoch ein Polynom, was sich perfekt für linear Filter eignet.
+Equirippel bedeutet, dass alle lokalen Maxima der Betragsfunktion gleich gross sind.
diff --git a/buch/papers/fm/00_modulation.tex b/buch/papers/fm/00_modulation.tex
index dc99b40..982d63c 100644
--- a/buch/papers/fm/00_modulation.tex
+++ b/buch/papers/fm/00_modulation.tex
@@ -3,11 +3,22 @@
 %
 % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
 %
+
+Durch die Modulation wird ein Nachrichtensignal \(m(t)\) auf ein Trägersignal (z.B. ein Sinus- oder Rechtecksignal) abgebildet (kombiniert).
+Durch dieses Auftragen vom Nachrichtensignal \(m(t)\) kann das modulierte Signal in einem gewünschten Frequenzbereich übertragen werden.
+Der ursprünglich Frequenzbereich des Nachrichtensignal \(m(t)\) erstreckt sich typischerweise von 0 Hz bis zur Bandbreite \(B_m\).
+Beim Empfänger wird dann durch Demodulation das ursprüngliche Nachrichtensignal \(m(t)\) so originalgetreu wie möglich zurückgewonnen.
+Beim Trägersignal \(x_c(t)\) handelt es sich um ein informationsloses Hilfssignal.
+Durch die Modulation mit dem Nachrichtensignal \(m(t)\) wird es zum modulierten zu übertragenden Signal.
+Für alle Erklärungen wird ein sinusförmiges Trägersignal benutzt, jedoch kann auch ein Rechtecksignal,
+welches Digital einfach umzusetzten ist, 
+genauso als Trägersignal genutzt werden kann.\cite{fm:NAT}
+
 \subsection{Modulationsarten\label{fm:section:modulation}}
 
 Das sinusförmige Trägersignal hat die übliche Form: 
 \(x_c(t) = A_c \cdot \cos(\omega_c(t)+\varphi)\).
-Wobei die konstanten Amplitude \(A_c\) und Phase \(\varphi\) vom Nachrichtensignal \(m(t)\) verändert wird.
+Wobei die konstanten Amplitude \(A_c\) und Phase \(\varphi\) vom Nachrichtensignal \(m(t)\) verändert werden können.
 Der Parameter \(\omega_c\), die Trägerkreisfrequenz bzw. die Trägerfrequenz \(f_c = \frac{\omega_c}{2\pi}\),
 steht nicht für die modulation zur verfügung, statt dessen kann durch ihn die Frequenzachse frei gewählt werden.
 \newblockpunct
@@ -18,10 +29,16 @@ 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}
+Um modulation zu Verstehen ist es am Anschaulichst mit der AM Amplitudenmodulation, 
+da Phasenmodulation und Frequenzmodulation den gleichen Parameter verändert vernachlässige ich die Phasenmodulation ganz.
+
 To do: Bilder jeder Modulationsart
 
 
diff --git a/buch/papers/fm/01_AM.tex b/buch/papers/fm/01_AM.tex
index 921fcf2..714b9a0 100644
--- a/buch/papers/fm/01_AM.tex
+++ b/buch/papers/fm/01_AM.tex
@@ -11,19 +11,61 @@ Nun zur Amplitudenmodulation verwenden wir das bevorzugte Trägersignal
 \[
     x_c(t) = A_c \cdot \cos(\omega_ct).
 \]
-Dies bringt den grossen Vorteil das, dass modulierend Signal sämtliche Anteile im Frequenzspektrum inanspruch nimmt 
-und das Trägersignal nur  zwei komplexe Schwingungen besitzt. 
+Dies bringt den grossen Vorteil das, dass modulierend Signal sämtliche Anteile im Frequenzspektrum in Anspruch nimmt 
+und das Trägersignal nur zwei komplexe Schwingungen besitzt. 
 Dies sieht man besonders in der Eulerischen Formel
 \[
     x_c(t) = \frac{A_c}{2} \cdot e^{j\omega_ct}\;+\;\frac{A_c}{2} \cdot e^{-j\omega_ct}.
+    \label{fm:eq:AM:euler}
 \]
-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.
-\newline
-\newline
+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.
+
+Dabei entseht wine Umhüllende kurve die unserem ursprünglichen signal \(m(t)\) entspricht.
+\[
+    x_c(t) = m(t) \cdot \cos(\omega_ct).
+\]
+
+\begin{figure}
+	\centering
+	\input{papers/fm/Python animation/m_t.pgf}
+	\caption{modulierende Signal \(m(t)\)}
+	\label{fig:bessel}
+\end{figure}
+%
 TODO:
+Bilder 
 Hier beschrieib ich was AmplitudenModulation ist und mache dan den link zu Frequenzmodulation inkl Formel \[\cos( \cos x)\]
 so wird beschrieben das daraus eigentlich \(x_c(t) = A_c \cdot \cos(\omega_i)\) wird und somit \(x_c(t) = A_c \cdot \cos(\omega_c + \frac{d \varphi(t)}{dt})\).
 Da \(\sin \) abgeleitet \(\cos \) ergibt, so wird aus dem \(m(t)\) ein \( \frac{d \varphi(t)}{dt}\)  in der momentan frequenz. \[ \Rightarrow \cos( \cos x) \]
+\subsection{Frequenzspektrum}
+Das Frequenzspektrum ist eine Darstellung von einem Signal im Frequenzbereich, das heisst man erkennt welche Frequenzen in einem Signal vorhanden sind.
+Dafür muss man eine Fouriertransformation vornehmen. 
+Wird aus dieser Gleichung \eqref{fm:eq:AM:euler}die Fouriertransformation vorggenommen, so erhält man
 
-\subsection{Frequenzspektrum}
\ No newline at end of file
+%
+%Ein Ziel der Modulation besteht darin, mehrere Nachrichtensignale von verschiedenen Sendern gleichzeitig
+%in verschiedenen Frequenzbereichen über den gleichen Kanal zu senden. Um dieses Frequenzmultiplexing
+%störungsfrei und mit eine Vielzahl von Teilnehmern durchführen zu können, muss die spektrale Beschaffen-
+%heit der modulierten Signale möglichst gut bekannt sein.
+%Dank des Modulationssatzes der Fouriertransformation lässt sich das Spektrum eines gewöhnlichen AM Si-
+%gnals sofort bestimmen:
+%A c μ
+%F
+%·(M n (ω−ω c ) + M n (ω+ω c )) (5.5)
+%A c ·(1+μm n (t))·cos(ω c t) ❝ s A c π (δ(ω−ω c ) + δ(ω+ω c )) +
+%2
+%Das zweiseitige Spektrum des Nachrichtensignals M (ω) wird mit dem Faktor A 2 c μ gewichtet und einmal
+%nach +ω c und einmal nach −ω c verschoben. Dies führt im Vergleich zum Basisbandsignal zu einer Verdop-
+%pelung der Bandbreite mit je einem Seitenband links und rechts der Trägerfrequenz. Weiter beinhaltet das
+%Amplitudendichtespektrum je eine Deltafunktion mit Gewicht A c π an den Stellen ±ω c , d.h. ein fester, nicht-
+%modulierter Amplitudenanteil bei der eigentlichen Trägerfrequenz.
+%Das Amplitudendichtespektrum ist im nachfolgenden Graphen für A c = 1 und μ = 100% dargestellt.5.3. Gewöhnliche Amplitudenmodulation
+%47
+%Abbildung 5.12: Amplitudendichtespektrum von gewöhnlicher AM
+%Für das Nachrichtensignal wurde in diesem Graph mit einem Keil symbolhaft ein Amplitudendichtespektrum
+%|M (ω)| gewählt, bei welchem der Anteil auf der positiven und jener auf der negativen Frequenzachse visuell
+%gut auseinandergehalten werden können. Ein solch geformtes Spektrum wird aber in der Praxis kaum je
+%auftreten: bei periodischen Testsignalen besteht das Nachrichtensignal aus einem Linienspektrum, bei einem
+%Energiesignal mit zufälligem Verlauf aus einem kontinuierlichen Spektrum, welches jedoch nicht auf diese
+%einfache Art geformt sein wird
\ No newline at end of file
diff --git a/buch/papers/fm/02_FM.tex b/buch/papers/fm/02_FM.tex
index fedfaaa..a01fb69 100644
--- a/buch/papers/fm/02_FM.tex
+++ b/buch/papers/fm/02_FM.tex
@@ -6,9 +6,65 @@
 \section{FM
 \label{fm:section:teil1}}
 \rhead{FM}
-\subsection{Frequenzspektrum}
-TODO
-Hier Beschreiben ich FM und FM im Frequenzspektrum.
+\subsection{Frequenzmodulation}
+(skript Nat ab Seite 60)
+Als weiterer Parameter, um ein sinusförmiges Trägersignal \(x_c = A_c \cdot \cos(\omega_c t + \varphi)\) zu modulieren,
+bietet sich neben der Amplitude \(A_c\) auch der Phasenwinkel \(\varphi\) oder die momentane Frequenzabweichung \(\frac{d\varphi}{dt}\) an.
+Bei der Phasenmodulation (Englisch: phase modulation, PM) erzeugt das Nachrichtensignal \(m(t)\) eine Phasenabweichung \(\varphi(t)\) des modulierten Trägersignals im Vergleich zum nicht-modulierten Träger. Sie ist pro-
+%portional zum Nachrichtensignal \(m(t)\) durch eine Skalierung mit der Phasenhubkonstanten (Englisch: phase deviation constant) 
+%k p [rad],
+%welche die Amplitude des Nachrichtensignals auf die Phasenabweichung des
+%modulierten Trägersignals abbildet: φ(t) = k p · m(t). Damit ergibt sich für das phasenmodulierte Trägersi-
+%gnal:
+%x PM (t) = A c · cos (ω c t + k p · m(t))
+%(5.16)
+%Die modulierte Phase φ(t) verändert dabei auch die Momentanfrequenz (Englisch: instantaneous frequency)
+%ω i
+%, welche wie folgt berechnet wird:
+%f i = 2π
+%ω i (t) = ω c +
+%d φ(t)
+%dt
+%(5.17)
+%Bei der Frequenzmodulation (Englisch: frequency modulation, FM) ist die Abweichung der momentanen
+%Kreisfrequenz ω i von der Trägerkreisfrequenz ω c proportional zum Nachrichtensignal m(t). Sie ergibt sich,
+%indem m(t) mit der (Kreis-)Frequenzhubkonstanten (Englisch: frequency deviation constant) k f [rad/s] ska-
+%liert wird: ω i (t) = ω c + k f · m(t). Diese sich zeitlich verändernde Abweichung von der Kreisfrequenz ω c
+%verursacht gleichzeitig auch Schwankungen der Phase φ(t), welche wie folgt berechnet wird:
+%φ(t) =
+%Z t
+%−∞
+%ω i (τ ) − ω c dτ =
+%Somit ergibt sich für das frequenzmodulierte Trägersignal:
+%
+%Z t
+%−∞
+%x FM (t) = A c · cos  ω c t + k f
+%k f · m(t) dτ
+%Z t
+%−∞
+%
+%m(τ ) dτ 
+%(5.18)
+%(5.19)
+%Die Phase φ(t) hat dabei einen kontinuierlichen Verlauf, d.h. das FM-modulierte Signal x FM (t) weist keine
+%Stellen auf, wo sich die Phase sprunghaft ändert. Aus diesem Grund spricht man bei frequenzmodulierten
+%Signalen – speziell auch bei digitalen FM-Signalen – von einer Modulation mit kontinuierlicher Phase (Eng-
+%lisch: continuous phase modulation).
+%Wie aus diesen Ausführungen hervorgeht, sind Phasenmodulation und Frequenzmodulation äquivalente Mo-
+%dulationsverfahren. Beide variieren sowohl die Phase φ wie auch die Momentanfrequenz ω i . Dadurch kann
+%man leider nicht – wie vielleicht erhofft – je mit einem eigenen Nachrichtensignal ein gemeinsames Trägersi-
+%gnal unabhängig PM- und FM-modulieren, ohne dass sich diese Modulationen für den Empfänger untrennbar
+%vermischen würden.
+%
+%Um die mathematische Behandlung der nicht-linearen Winkelmodulation etwas zu verkürzen, ist es aufgrund
+%dieser Äquivalenzen gerechtfertigt, dass PM und FM gemeinsam behandelt werden. Jeweils vor der Modu-
+%lation bzw. nach der Demodulation kann dann noch eine Differentiation oder Integration durchgeführt wird,
+%um von der einen Modulationsart zur anderen zu gelangen.
+%\subsection{Frequenzbereich}
+%Nun
+%TODO
+%Hier Beschreiben ich FM und FM im Frequenzspektrum.
 %Sed ut perspiciatis unde omnis iste natus error sit voluptatem
 %accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
 %quae ab illo inventore veritatis et quasi architecto beatae vitae
diff --git a/buch/papers/fm/03_bessel.tex b/buch/papers/fm/03_bessel.tex
index 760cdc4..3c2cb71 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 +67,122 @@ 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*}
-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),
+%intertext{}   Funktioniert nicht.
+wobei mit dem Additionstheorem \eqref{fm:eq:addth2} \(A = \omega_c t\) und \(B = 2k\omega_m t \) ersetzt wurden.
+Nun kann die Summe in zwei Summen 
+\begin{align*}
+    c(t)
+    &=
+    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) \}
+    \\
+    &=
+    \sum_{k=\infty}^{1} J_{2k}(\beta)  \underbrace{\cos((\omega_c - 2k \omega_m) t)} 
+    \,+\,J_0(\beta)\cdot \cos(\omega_c t) 
+    \,+\, \sum_{k=1}^\infty J_{2k}(\beta)\cos((\omega_c + 2k \omega_m) t) 
+\end{align*}
+aufgeteilt werden.
+Wenn bei der ersten Summe noch \(k\) von \(-\infty \to -1\) läuft, wird diese summe zu \(\sum_{k=-1}^{-\infty} J_{-2k}(\beta)  {\cos((\omega_c + 2k \omega_m) t)} \)
+Zudem kann die Besselindentität \eqref{fm:eq:besselid3} gebraucht werden.  \(n \) wird  mit \(2k\) ersetzt, da dies immer gerade ist so gilt: \(J_{-n}(\beta) = J_n(\beta)\)
+Somit bekommt man zwei gleiche Summen
+\begin{align*}
+    c(t)
+    &=
+    \sum_{k=-\infty}^{-1} J_{2k}(\beta) \cos((\omega_c + 2k \omega_m) t)
+    \,+\,J_0(\beta)\cdot \cos(\omega_c t+ 2 \cdot 0 \omega_m) 
+    \,+\, \sum_{k=1}^\infty J_{2k}(\beta)\cos((\omega_c + 2k \omega_m) t) 
+\end{align*}
+Diese können wir vereinfachter schreiben,
+\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*}
+da \(2k\) für alle negativen, wie positiven geraden Zahlen zählt.
+%----------------------------------------------------------------------------------------------------------------
 \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 
 \[
-     \sum_{n\, \text{ungerade}} -1 \cdot J_{n}(\beta) \cos((\omega_c + n\omega_m) t).
-     \label{fm:eq:ungerade}
+    s(t)
+    =
+    \sum_{n\, \text{ungerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t).
+    \label{fm:eq:ungerade}
 \]
-Substituiert man nun noch \(n \text{mit} -n \) so fällt das \(-1\) weg.
-
+, mit allen positiven und negativen Ganzzahlen 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 
 \[
     \sum_{n\, \text{ungerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t)
 \]
@@ -130,17 +190,16 @@ ergeben zusammen
 \[
     \cos(\omega_ct+\beta\sin(\omega_mt))
     =
-    \sum_{k= -\infty}^\infty J_{k}(\beta) \cos((\omega_c+k\omega_m)t).
+    \sum_{k= -\infty}^\infty J_{n}(\beta) \cos((\omega_c+ n\omega_m)t).
 \]
 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 +210,7 @@ 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..4074765 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": [
@@ -11,11 +11,12 @@
     "from scipy.fft import fft, ifft, fftfreq\n",
     "import scipy.special as sc\n",
     "import scipy.fftpack\n",
+    "import matplotlib.pyplot as plt\n",
     "import matplotlib as mpl\n",
     "# Use the pgf backend (must be set before pyplot imported)\n",
-    "#mpl.use('pgf')\n",
-    "import matplotlib.pyplot as plt\n",
-    "from matplotlib.widgets import Slider\n",
+    "# mpl.use('pgf')\n",
+    "\n",
+    "\n",
     "def fm(beta):\n",
     "    # Number of samplepoints\n",
     "    N = 600\n",
@@ -27,7 +28,7 @@
     "    #beta = 1.0\n",
     "    y_old = np.sin(fc * 2.0*np.pi*x+beta*np.sin(fm * 2.0*np.pi*x))\n",
     "    y = 0*x;\n",
-    "    xf = fftfreq(N, 1 / 400)\n",
+    "    xf = fftfreq(N, 1 / N)\n",
     "    for k in range (-4, 4):\n",
     "        y = sc.jv(k,beta)*np.sin((fc+k*fm) * 2.0*np.pi*x)\n",
     "        yf = fft(y)/(fc*np.pi)\n",
@@ -42,12 +43,12 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 114,
+   "execution_count": 6,
    "metadata": {},
    "outputs": [
     {
      "data": {
-      "image/png": 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",
+      "image/png": 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",
       "text/plain": [
        "<Figure size 432x288 with 1 Axes>"
       ]
@@ -56,31 +57,10 @@
       "needs_background": "light"
      },
      "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "# Number of samplepoints\n",
-    "N = 800\n",
-    "# sample spacing\n",
-    "T = 1.0 / 1000.0\n",
-    "x = np.linspace(0.01, N*T, N)\n",
-    "\n",
-    "y_old = np.sin(100* 2.0*np.pi*x+1*np.sin(15* 2.0*np.pi*x))\n",
-    "yf_old = fft(y_old)/(100*np.pi)\n",
-    "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>"
       ]
@@ -92,17 +72,37 @@
     }
    ],
    "source": [
-    "fm(1)"
+    "# Number of samplepoints\n",
+    "N = 800\n",
+    "# sample spacing\n",
+    "T = 1.0 / N\n",
+    "x = np.linspace(0.01, N*T, N)\n",
+    "\n",
+    "y_old = np.sin(100* 2.0*np.pi*x+1*np.sin(15* 2.0*np.pi*x))\n",
+    "yf_old = fft(y_old)/(100*np.pi)\n",
+    "xf = fftfreq(N, 1 / N)\n",
+    "plt.plot(xf, np.abs(yf_old))\n",
+    "#plt.xlim(-150, 150)\n",
+    "plt.show()\n",
+    "\n",
+    "fm(2)"
    ]
   },
   {
    "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 +111,6 @@
       "needs_background": "light"
      },
      "output_type": "display_data"
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "0.7651976865579666\n"
-     ]
     }
    ],
    "source": [
@@ -133,19 +126,19 @@
     "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))"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 85,
+   "execution_count": 131,
    "metadata": {},
    "outputs": [
     {
      "data": {
-      "image/png": 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Dhw7gKJ+OSqXC1atXOX78OBMTE9tu2y0ouhu8KKJsN+8Xi0WGh4dJp9OUSiXm5+e5desWiUSiQ4AH7bZ142UhvL1CCLGpk6a7BjAMw04pUaFQeGop0W5c2vHx8YP6Gi8EnwrC01qzsrJCKpU6kP4/rTUzMzM8evSI4eHh50Z2s7OzzMzM8Oabb+7oSboVcXW3C73sN/XG3tVms0mxWOy0bnXH/w5SvfhVODe7wVaWdLuUaHp6Gq11jwzWxkTbbrK0fQvvBaPtwt65c4fz58+TSqX2tV4QBFy7dg3Lsrhw4QLz8/MHdKTr2HjDdTf+X7x4cceZ373etC+rK5xMJpmcnOy0btVqNYrFYsdq6e5c2E92/GXpcHhWv0G3zBX01gA+ePAAIUSPDJZSqu/SvuzYmJhoV67vB+VymWvXrnXcyUqlcqDN/sAm66u78X9qampXJPayEtdBoNttO3r0KFEUdW7aR48e9dy0uxVA+LRZeE/DVjWAa2trLC0tcffuXZrNJrOzswwPD28bS20PTXqV8UoS3lbtYfvpU21nRBcWFnj77bc7VuJBq5t0ryml7PTf7rV4eT9Z2heJvRCOYRgMDg4yODgIxDdtqVTakwDC85Zl2govknQ36gB+7WtfI5VKPbUGsG/hvQCEYbhlbZ1hGHsqdfB9n6tXr5JMJrl06VLP0+1ZEJ4QojPIx3Xdp/bfPm2tT6uF9zRYlrVpfm1b/qpdINsmwI1lGy+DhfcyHEMbQohOLzVsXQP4W7/1W3sqPP7Sl77EH/gDf+AWYAA/pbX+sSccw38A/FPgotb66/v7Rk/GK0N4T6utMwxj1+TUzmqdOnWqEzjvxrMgPK01H3744RMlpHaDzzLhbUQikeDQoUOdurV6vd6RzvI8ryf+9zKQzcsiHLAVNtYA1mo1vvSlL3Hjxg3+yB/5I7z33nt8//d/P1/4whe2XSeKIv78n//zAH8AmAU+EEL8otb6evd2Qogs8EPAV5/RV+rglSC8nbSH7UadWGvNgwcPWF5efmJRb3vNgyS85eVlKpUKb7zxxoGk919lwnuWhNMtgNAu26hUKhSLRWZmZgiCAMuyKBaLL0wA4WUmvG60Y6l/6S/9JX7lV36Ff/2v/zUPHjyg0Wg89bNf+9rXOHXqFPfu3bvfWuv/BP4gcH3Dpj8K/Djwlw76+DfipSa8J0mvb4WdurS+73PlyhUymcy2Rb3t/R0E4XU3/rczYweFrQjPdd2OTPduPve88Lz33S2AALF24erqamd4j2maHff3eQkgvCqE141ms0kmk+Gdd97Z0fZzc3McPny4+6VZ4Ju6XxBCvAsc1lr/KyHEZ5fwtpNe3wo7IadisciNGze27Evd65pPw8bG/08++eTAbngpZc9abct1fn4erTWO43QC/c+ykPdVg5SSdDrNsWPHgFgAoVgs9gggdMf/nsV5e1kIbzfX4kFL0gshJPA/Aj94YIs+BS8l4e2lPWw7C09rzf3791ldXeW9997b8RSvjYSyW2zV+H/QbnL7+MIw5MqVKyQSCd577z2EEJ1Afrv/slvI87OMjTE8x3F6BFC3EkBon7eDEkB4WQhvpxlrrfWu74XJyUlmZma6X5oC5rr+nQVeB36jdQzjwC8KIb7nWSUuXirC26tuHdCR2NkIz/O4fPkyhUKB999//7lcZNs1/u+XRLvRXqtWq3H58uVO/WD7YbGxkLdarbK6usrc3ByNRoN79+51hDyf9833sspDPUkAoVgsMjc31xFAaLdt7dXieRlKY9rHsZvvsJtjvnjxInfu3EEIcZyY6P448B+239dal4HhrrV/A/jPPxNZ2t1Ir2+FrbK0q6ur3Lx5k7Nnz3b025412o3/iURiyxjhXvtfn4R24fIbb7zRUXF5UstZLpcjl8tx/PhxvvrVr5LL5VheXubOnTvP1f190YmW3WRpu9u2jh8/3hFAaFvOexXufFm6PXbaVraX38w0Tf723/7bfPd3f/cvE5el/AOt9TUhxF8Dvq61/sVdL7pPvBSE17bq9qNw0p2lVUpx7969HnXg54GdNP4fVGZVKcXMzAz1ep3f83t+z65r+aSUPcWnG0cQtt24l0nH7qCwn7KUtgBCu2uhPby7XbS7UwGEKIpeKcLzPG9PMlRf+MIX0Fqf6X5Na/1Xt9pWa/1tu97BLvFCCW8/LuxGtGN4ruty+fJlBgcHe9SBnzV22vh/EDG8tpveDq4fxGCdjbVXbff36tWrPXMsCoXCS3Gj7gcHWYfXLYCgu2bXtuOmTxJAeFksvN0M8Nlvn/rLgBdGePuRXt8KhmFQr9f5xje+wfnz5zstSM8a3Y3/O9HL269Lu7a2xrVr1zhz5gyJRIKHDx/uea0nYaP7255jsbKywt27dzttXIODg6TT6V3/ds+y8Le6sMytf/NbrD1eRGjN8MljFA6PM/nu65gtwnlWZCPE5tm1G2Wb8vk8g4ODhGH4UhDeZ0ntGF4A4e2mtm6nUEoxPT1NtVrlW77lW56bC7aXxv+9Ji201szOzjI7O8s777xDKpWiWq0+l3jYxjkWbSvm4cOHL4X7uzb7mN/8H36Ku7/9NRqrJVQQIqQkdH2UEAggYZsUJsc4+v7rnPqP/ij5yWev69b94NgogLC0tNTJfA4ODpLL5V4IAX6WxD/hOROe1ppisdgZBH0QT/lms9nJwubz+QO/4doW2caLca+N/3shvLYVCfRYkS+q02Ir97dYLG5yf591F8PC1Vv80o/8D8xdu0nY9ImUwjIk9VqTfC5D1fMZyKbQSuNV6lTlEh//3C9z55d/i8MX3+D/8nf+BonC1kPTnwW6BRBSqRSe55FKpVhcXOT27dudpv29Ws57wWdpYhk8R8Jr19ZNT08zPj5+IImExcVF7t69y2uvvUYqleLq1asHcKS96FY3gfh73Lp1a8+N/7t1aZvNJp988gmHDh3aNPv2ZWgt67Zijh07duDu71bQWvOv/rO/xuV/+WsoPyRsNHHDiGw2TbNSJWcahCqikM9QKtcoDORAKRCCRNKhulrm8Vc/4R998x/kW37k/8WF/9sfOYAzsTsopTYJIGy0nNtzKwYHB3dcO7pb7Mal7RPeDrCVbt1+B7i0SafZbHLx4kVs2+50ZRw0upMMbWtydHR0z43/u7Hw2kO3L1y40GmL6sbLQHgbsdH9bRc/d7u/zWaTMAz3tP7K7fv8s//HF1m4N40EhCHxIkUCjed6GIkETc/HCUPMhM2QY7BaqmBIQUZHWNk0CJCGxK01+fW/9NdZ+q1/x7f+3R97rj21SqlNIqZbNe2XSiVu3ryJ7/s9AggHkahqH8dOXOm+S7sDbFVbt1cZpzYajQaXL1/epDay33WfhDbhLS8vc/v2bV577bV9dSrsxMLr7gzZrqxmP4T3vIhyo4pJtVrl5s2b3L17FyHErtzfy//4n/Pr/83fpFmpkXBsdKRw/YCBbBItJZW1KpZjY5oG1aaPY1s0hMFgxgQhKddd7FKVoaE81WIFhGBsOM+9X/5Nyt/8PXzHz/2vpA8/Hzn/pxUebzW3oh3/m5mZQWvdI4C6V7Le6UzaPuE9Be3ExMbaOsMw9vx0b8fNLly4QD7fG3s5yA6GbgghuH//PvV6/UBq+p5GeO0WsWQy+dTOkJfRwtsObfe3rWKSSCQ67u+9e/ewLOuJ7u/X/se/x1f/l3+M7/qg4u+8VqqSzKbwIkWz2iRnG+BYrFUaDA/nWFmpUMglUVrjiIhCIU3kB6yuVhgaymE2G5SLFVIJk2a1zq/+4f87/97f+W8ZvPjuMz8Xu+1wkFJukm0vlUqbBBAGBwd3Nbi779IeANpuLGzOwu7FpY2iqGPW70cwc7fwfZ9yuYzjOAdW07edjNXGFrGn4VUjvI3Yifs7ODjIrf/5Z7jyT/4lbsNFRwqZsCmXa6RyKbTn0Yw06UyKRtNFNppMDCSYXakwMpyltlYnUprceAHVaLDmBmRSFpHrksylaSxXSJiSUEVUllb5nT/7w3zTj/8Io9/5+57pd99va5lpmj2F420BhJmZGarVKul0uhP/265geDdZ2qmpqT0f78uCZ0J420muG4aB7/s7Xqtd+jE5ObkpaP8s0a6dymQyB7rf9qzbjWhbr90tYk/DkwjvZejR3A5PqsPbyv397R/5cR596TeRYUQqYcdhkqbH5MQA5apLNYJsJkG93sA0JMlkgqVig2MjKeaLNbSGQ6M5VpbKmJbB2IBDvRHihYowiJgYS1NZdaEZMjScRTUaXP6Rv8HbQjL8Hd/2TM/BQZahbCWAUCwWO8rauVyuYyF2u7A7tTT7Wdo9YjextsePH/Pw4cM9z3zYC7TWPHz4kKWlJd59910ePHhwoL2vG0lKKcXt27dpNBq7tl5fdQtvOwghuPm3/jdWfvMrmIZBqBRBGFKrNkmnHRqlKqZhMDk1yPxcEcs0sA1Jpdxg6ugwSzOrHJrII4Vmcb4KwNhoFq8ZIGSEk7RJGbCy3GRwIIFh2wQ1l0TSoLZa5vaP/09YAzny7z8b9/ZZqqV0CyAcPnx4kwBCu3RoYGBgxwXQbYv7VcczI7wnWRmmaT41hhdFETdu3CCKIi5durSvkXy7wVaN/wfd7N8da2y3iA0ODvLOO+/s2jL7NBPezb//f3D7//xFfC8gROOFiqYfUBjO0aw2iDQkpaKyUOLQRI7SWpNazePI0SEWZlaRUmBbJkEY/3ZHjg6yPFsCYOzYCM3lMs1GfB1amTSUqwRBRCVUjI+maTya59Zf/mtc+Hs/Sfr4sQP/fs9THmqjAEK7dKhYLLKyskKj0WB4eLgjgLrVcX0aJpbBS2jh1Wo1rly5wuHDh5mcnHxubUvtxv8TJ070yK/vRjp+J2gTaLtFbD9KLvshvJfZ7Z39pV/jo7/903hhBAKiMMDzPAqDWWo1F6RB0jFo1lwSaYeg6jKSd4jyDgszRUxTMjScZXmuRCJhcvzkKAsPlwEYnhyk9rjI4NQwyw+XOXRqnPKDRXIjOfBqDA4ncRsBCcfEXSly74f/Mq/9w5/CPOCb/UXq4XXHThuNBidOnKBer/cIILT1/9oCCJ+WLO1zP+NPIrx269Tly5d5/fXXdz2jtb32Xqyx2dlZrl27xptvvrlp1oRhGAdqRQkhWFtb4+bNm7zzzjv7kq16VS287R5Kq1eu88H/+ycwogArCpFhSFIYHD40iIhCJGCbAq/hYyctJNCo+yQSCSzLxk6YDA5lKS1VMC2D/FCW0ItjpuPHRqnNl4hCTWmuyJGz41QeLiGEoFFuMHFqHK/UJGj6yIyDaVt4K2Xu/vB/ceDn+WURAI2iiGQyyfj4OOfPn+fixYucOnUKgPv37/PVr36VP/2n/zQrKyu4rrurtb/0pS9x9uxZhBB3hRBf3Pi+EOI/E0JcF0JcFkL8qhDi6MF8qyfjmZ3x3bi07VKMUqnEpUuX9hwr2K01FkURV65coVgscunSpS2fYAepUBxFEY8ePcLzPC5evLhv9YntCG8nN+jebmIR/2mVGj3pD13/R7Q+04J8QpDcrzf48D//UZTrEoSKSCvMhEUyl2JtuYwjDcaG0gRBhGVLDA2Nqsuhw4Msz6yimh7jEwOsLVdIZBIMD2Uoz69RXihz+u2jlB4tAwIrYTE0miesxINonJTF8HCWyvQyTj6FlXIwIrBTFn6phvfoIY//57+5h3P1ZLwshLfxONoCCFNTU7zxxhtcunSJH/zBH6RUKvHDP/zDvP/++/zUT/3UU9dtTyz7pV/6JYDXgO8XQry2YbOPgPe11m8Sj2j87w7siz0BL9ylbbuSR48eZXJy8kDX3g7t7O/TXOeDcmnbBdPtOQkHUdW/1TGHYdhR5hgaGtokS/SElVp8JNoLb7+fHWQYtVKIJ2zzxhtvtpZpE65Ga/joP/nL1FeLhGiwTWzbIFKa0kKJ7HAWv+GhqhFHjg6zML9Gs+oydXKcpQeLZAYyGAJkEJEbSmNrWFuqIA3J1IlRVm/NkxlII6QgaUrqi2sAjJwZR69UcVcqACQGshjVGlG5QcM1yUwN4ZdrlH79t0i/9z6Fb/6Wp5zLneFlIbynZYullHz+85/HNE1+6Zd+CSklq6urT123PbHsxIkTaK39rSaWaa1/vesjXwH+5N6/yc7wwghPa83MzAxzc3NP1ZDbKXZqje2m8f8gLLx2l8aFCxeAeJrTs0Cj0eCTTz5hcnKSVCrVKa0xDIPR0THyhQKJZAIhJG+/+x7S3JwR3lkMNCaqrc6Lbv1HoxFPOG+u6+LYNqKVFALBzR//n1i6fpsgilCAkOA1POqVJoWRHPVyA9s2UWGEu1xhcDgHQxmWHiySH0yjg5B6zaMwkuXQWI6HNxewHJPRiQLL95cAmDw9RHlmhWalCcDQ0VHkWp2w9UBLD6VRxTKp0QKNiouZcjBESKQ0Qc1l+R/8XdIXXsfK7V9w4GUhvJ3Cdd3Og3onnslOJpZtwJ8Gfml/R/l0PPcsbZtAPvnkEyzL2pGG3E7xNAtvL43/+7Hw2i1ixWKx06VRqVQOfLg3rPfdvv7GG6RSKZTS5AoFjh4/0fktum+yJ7UTxZJFbUqL/x7/bf3/Ao1Q0SZLsHehJ18DdiKJbu1LClj5rd9l5he/RNIU2IZFGAa4QlJu+AxNDbO2UCKVTRI0fZyEQ6NcZ2JigMePSwyMFQhqLl7DY/jICN5qhVLNZfjoKEakKM4UARg7fYjq/QVyh4ZYqS8xfvYQzYeLhEozfGaC0FeolTKh69NYLpMYzSOIqC9WSU/ksZI2hBELP/HfcPi//u939dtshVeN8OBgpNy2ghDiTwLvA7/3meygC8/9jJfLZer1OmNjY1y4cOFAG7a3I7xms8kHH3xAIpHg7bff3nG9214tvCAI+OijjwjDkPfee6/jVh5koqE9inFpZYUgirj4TZ8jlc6AkEjDQAjZQzrdf48iRbPZpFyuUKnVaLheXIyrNaGGSEOkBYr4j2796Y7hbToepTt/lNYopbb8U61WW5L+Crdc4e6P/ySmDol8n7Dp4tdcLNfn+OkJio9XyQ5m8BseyUyCRrnO+JFhVu4uMDU1iFdu4DU8Dp2dxF0qEbo+6YEsY8NZKvMx2U29dYLagwVUqEDB2GuHadyb77SoCSGxfR/lxgXxZjpJZiSFasb/VoZFFPi4KxW8hQUW/vn/d99hjmcpgnrQeEYTywAQQnw78F8C36O13jyF64Dx3Fza9iSvhYUFUqnUjlqndosnZWn30/i/F8KrVqtcuXJlU4nLXtfrhtYaDTGhaM0bb8dDkbe7ddoXbKQ0WoiWdQXSTpCwe2WHatUqmWxu0+d7968RrT12W350HYkB6Cfc0Ml0pvUZxZ0f+a/xS2txkiJUhJHCySQxbIvKvXlOn5vkwZ15sgNZqstlxo+PsfpgkcGjI9RmVhk/NgKGweqdObTSDB8bJSjWWLr6iJHjYySzaYrXHiIAaRpkChkkCrd1bONvnqB67QEDpw7RWCqTOz4BlSq12TWsTAJpSozIJzE+iLuwgmFZuP/mX3A5MwhDo3vqX22fxxdNeLslsb1MLHvw4AEnTpyw2TCxrLXeO8D/Avz7WuulXR3MHvFcXNrugt5Lly7xla985Znsc6P7qbXm7t27+xrms1uCmp+f58GDB0+MS+7FwlNKoRGoFtl1rbb1B7QmarmmkQbVsspk12e2una11iSSSVzXxQ8ClNaYlo1pWcQzk+mss93Fr7VGi/g4Wq+s/1fHCaNEMsnqP/lZKldvEmhQaLAMEimLyAuozJUZODpK7dESh89MMnN9hvHjo6w+WGT4xDj1uVWiIGQ8n+HeJ/fQSjN+dpLao2WiIMS0TcbGB3j4wW0AnFyawnCOys2HJEcKGI7F6JkjVK8/iEuF7s0z8u5pmvdnIYyvoeTkYXSzQlht0JhbIX14GL9YIjU5yuErv036z/0Ia2tlpqenOxLobQJ8XoOj9oOdutV7Ief2xLLv+q7vArjB1hPL/nsgA/xsa/1prfX37PJr7O64nuXisD6D4dSpU4yNjQFPVhHeL7otvI3zaPf6NN0p4XXHB7frDtnpelprlKaL5LYnSaU0CghVe8stXM6uV4MwRLZKRuL9gEYgMRCWiWM9+YYNohDLMDuLxjYfndifRoDULRd487Ek0hmaj+d5/I9/ljAKY4vRMjClxKs08WtNBo+PU3m0SH5yCHe5xOSpMZbuLjJy+hDVR0uoMGLy9aOsXrnP4deOEiFYvf4ItCY5kCU/kGHlozsMnzmC13Cxw5D69AIAoRtw+L1zrHztWueYRt49h6GjDtk5IwNEtSpmOgHVBsIysXIZVK2Kt7yGlRBEX/5ZJr7nBzr9q7VajWKx2MmSH8T82meJnU5Oazabeyqh+sIXvsAXvvAFgJPt17onlmmtv33Xi+4Tz4zwtNY8ePCApaWlzgyGNtqxtoMmvLaF185OnjlzpqMmsZ81n0ZQnufxySefMDw8/FRh0KfVzmmIXc+nHJfWChBUag3sxLoaxtb7bmdVNQrdIjeDSLOJR7tJUWvd8/eo5RaHQQCJ7ktnq3geIDdkPFrE6Loei//dj2ObYCZNImFQaXjUVyoIKSkcHWXt3jyDx8cpP1xg+NQhlGXB2UnK9xfQkWLywlGKN6YByOWz3P/gBmhN/vAYVuBTa5FbYWSA5St38BpxZjZzeBwnivBmFkHGtYIj75yjcfMuAOnjU/FMDOURrFVwho8irSrJsQHqD2ZIjg5iJCyEYdD48Gsk3voc5tGzPfp17fkV3fNrN8o3vQzYjTTUp2FiGTxDwgvDEKXUlsOo24R30DJPUkqWlpbwfZ933313T3M0t1pzO8Jrk+tOW8S2Wq8dX9uWVrWm6bqEYYSVSHWsvm6yk6KbW2I2i5QmaFlvwBPdUa1jxzfScRZWQ+t4uq00gZDgOOtxv0hFBJ5HGIYYhhm7v4aBkCImvR7Ea639/M/iz82gwxAihfADMlqhBjJI02Dt3jxDJw+xdu8xI2enKN+ZY+DcEYLVCmjNodePUbzxCCEFE2+epvj160y9eQa34eLPLeI24o6AifcuUP3oOpmpCcr3Zxh4/SxMzxI2XUJg6P23EL7bITsAe2gAd/oRUauroP5wlqF3z9G4/yD+BpaFYYK/skby8ATur/wsqT/1w0irN+u9cX5tW76p7f56nsf8/PwLdX93qpTyaWkrg2dIeLZtc/LkyS3fexbqxEEQMDs7i2EYW5LsXvGkREi7jvDx48e7ItduC6+deFB6+6RDGEXUGk2cRArT3Ma5bbmXftRNnr0ry3gzwjBonSNBpCHsWtQSbAryxdZefKDt440tRYnhJDG67lkN1Gs1nFQKCUghaRuT7uwsjV/+N6hQo8MI5YeoSGGnEwxaBjPX5hg6M8na7TlGzk1Rvj3H4Lkj1O7MMn7uKM0JRenGQwzbZOT0EUpX7gCQTtqUrt1FhxHCkEy89waVD68AkBoo4HzTMO6Hlztnwzk0hmUKatfvd4479+4bNO7dJTF1CPfRDNKxyZw6guqS8zIyKaSpUI06IPAXl7B/++exf9/3bfMLbpZv+spXvoLv+y/U/d2NFt6nQTgAnnEM70nu204UU3aDdrdGeyj1QbrKW6mlRFHEtWvXkFJy8eLFXSvXJlMpgkj1ENems6QhiFSLiAROYmuXolatksxk8SONKWMLbWtoJMRlJ0qgMHiSSSlE22WO0SbE9tISMLtOcdv1bVuYSoOdTKKQ8S66jqn4d3+SqN6AMEJrgXQsrIRNWHcJS1UOvXuShQ/vMnr+MGu3ZjtkZ9gmjpCszM5jZ9MUxgYo34ytrrH3X6fyjesMv3aG0v1phk8c7pCdtG2shEU0/bhDdvL4YWSzSv3addKvnaV+/RaFS2/TuHEt1nJMpjDSKVLHDuE/nsUHEocncQZyBI8fIcfH0FKiAw8jnyOYeYA5cxN5+NyTTv4mmKbJ0aNHd+T+Pqtsbp/wnhMOysLTWjM3N8fMzAxvvfUWjUaDUql0AEe4jo0uaLubYWpqamMl+VOPtW3NnXvtwpZWWjtR4T+BtQzRcjlFnJzwI42RzHS2j1oJi/b9IVuuaag0fusr2LJTftYD0SItBQQhhNvWFOuOpai6ybBr3aQhCVV8zKAJwojqv/g5/Adz6AgQEtMxQGv8YqxXlzoyiltcY+S1o6zdnGbo/FGqt2cwUwnSY0NUbz7k0Lvnqc4sUn0whzRNRt44S+WjuFspkUxQGB6gdj3OzNpDg6TGBmlevU7m7TfxFpYZ+KZ3qF+/1uF6ISH/7us0b17vEIu/tETmtVN4d293vo81No7/IN7GW1gk9/YbqMVpQsvCTCeIPvlN5KFTYDz9ltqYsHua+/ussr87TRy+rEO4hRB/GPivNrz8JvDdWustuzZeWcJrW1lCiE63hud5B+4qd7u03S1iW00RexK01oRq+1yr19Jt24qMIHYhA6Xxtvl6GjBFXOLhhqpVjtKL9vpRGGHIuOg2VL0ubWLDQ1+0OiJi8uy19p60XaREq3g5Pnp/tUj0u7+GnTXxlEI3faJQEZRrmJkkdi5J/cE8+fPHqJSaDL12nOqtR1jZNMlClsbDxySnRpFuk8b8MlY2Q35yjMrlGwBkL5wheDyLMOJYWvr0CUSjivcoTmx40zMMfO4dGteudc6KPTqCJEC76/WuiSOHkcLHsNZvjfTrFwjuXcOeOES4tIA9OQXNRvyAajYRziHCxQWMD38ZefG7n/wDtX+DpxDNRvf3WWV/X/Uh3Frrnwd+vv1vIcSfBf4E8MtP+swLcWn3S3jdjf/dOvt7lYfaDlJKwjDk7t27lEqlzljInaBtsUXb1N3F3Q3x340N/CRoxb1CRaTX42/dLo7WGlPGpSWNQOGYAl9151rXYYj1BEmgBaHa2ozTWmG0CpQ3kmH3sdFl6QWq153OmK1ja+0i+id/DxmFCNOAvE2j2iSoNHCG8gjLoDG7Qu7cUWp3Z8icPcbi717HHszjJByas4tkTh0mWCvTWC0y9s1v480uUr8Tu7QD3/Q2jes30VFE5u03SYyP4t65A62wiT06gj2QxnDWSSZ97jRUi4QL86QvXMCbfUz6wnmi5TlU4NNo1DFyOZLHjhBO34ld3VQGozCAiUe4UsI+cgwzmSAqr2GkHMTKHBQfw+D2k892U5K12+zvbtzfV53wuiGEOAP8VeD3aL05XdbGC7Hw9hPD267x/6DFOiG+KKrVKoVCgffee2/HF6rqsuo2olqtYifTm4gkUjpWGUGjpYG7YQNFHDtTukVeQLnpIc11At64R6NVAOxFmmbHrRWbSLPtevqRRiuB3uprtiy4uG1MEmwsaWmtI4jd61ALQg3hV3+DaPoRWim0ihBhSHKsgJlKEnk+/lKJ7OnD1O/NkjlzlPrdR+TPHScsVnAXVsi9form9Bw6CEidPIqdSlB5vAhSMvS5t6l+fKX1ZQ3sjE3tyjVoCwKcP0NUXMKfnSNxLL7c5blTREuziNaDqHHnDrmL7+DdWXdrCUMyr7+Gd/2jzmvB/BzpU8dQC7HVqIIQrSoQesjRcVS5iHnnq+hLf2jbPuP9DPA5SPf30xLDE0JYwP8B/EWt9fR2274yLu1OGv8POvvbbhGzLIuzZ8/u6DNt9/VJNp0faWRiM9lprWnUa9hOkpCtA22twhCCSFNvvd9NdhCTpilAaYUbbp3EkAKCKOqUqPiRptm1nWPErWNa69Y2mjACN1r/Xrn2vdQiOa3BizReawPTiV8UfgP1q7+AjlRcmKzBsC2sZAJ3uUpYb5A+NkH9/izZs8eo331EcmIEmc2zcvsRhfcuULt5B5Qi/foZvIfTBEtL2GMjpCZHOmRnFfKkjo1Tv3yZ3JvnqHx0jfzFt2lcX3dh/YVFBn/vJZqXL6+fi0SC9NmTGAkDv0VCwrLInD9DOH0XYVoQhch0hsTEMNK2O/E/I5VCphLo5Tm0NNCGia7XkA8/Rh1/5wlXwMEKB+zH/d1qGPhWqNfr+xKqfQ74UeCa1vr/97QNXxjh7WZyWbPZ5PLly4yOjm5b2HuQYp3tAUJvvvkml7tukO3Qtuq2OrpIadxo6/e8ZpNAC6QT67VtJDuj9VIj0NgGLZeVTdtIAXVf0ZIq2bSNII7xRVFEM9AtvbretdpuuKkVfqQJtjidhtCxdaQ07obni0BjyVYcD+DnfxpdqbWsQwEJB4EmXIsVSSI/Q+Ph45js7k2TmBpHuw2i8iqFb3qT2uU4IVG4+Da1K1fjfYwMYQ4UaFy+BUDy5FFk5OK2auUM2yD/zgWaXWTnTB7CSplYpqDZes0eH8dMWfgP7mKNjMafzeVITgwTztyLP3f6LMHCYxKjg+jSMn6lhJHKYE8dgaUZFGNgWoh6ET0c94eL4iyMnYLU1kK2Bz2xrI2duL8DAwMMDQ2RyWQ+FRaeEOLbgP8A2NG0pWcew9typ7twaXfT+H8QFt5OW8Q2fW6bxEQzUB2rQANuo0EilUJoTaXpYlhOR7YmaFmHgth9DSKodvmOfsvKEi3X0W3UsBMpKl0mY8oUPe6mJSDUioqnUBqylugR54zjgPH/66GiFkDS7FVaMUTszvqRphpobCk7mV+hNaYRu7GNUNMEbBP0/WtEN66io1hBBSkhjAirVcxUCiOZpPjhnXXL7sghomoZ1XTJvvEalYdLICX5997okF3bRbUTJkUg9dZ5oumHqNbvbk6MQ7NGsLzWOfbsW68TzE8TrvioZh1tGGTOniZamidaibP6wfIS6bfegtIi0dLj9ZMXBDhDBfTaSvzvKMQ6chI9fy8uWVpdxD7/FmJtBrU8h546DpUixqOPiM5/69bXynOShnqa+9ueH53NZkkkEk9c52WdWCaEGAD+N+A/1FpXd/KZl9albTf+l8vlHScK9kt4ruty+fJlRkZGntoi1n2cUcsq2vAGvooJbCM8z0VYNhESY0PPqiYmqEaoaWweX4smzqJGGqp+hDKThBv8YyHAbGnZVf1ok5UWqNhNlcSxpHqkCDfsK66za5FcqKlu8I0FGlvGSY1mpONSk9b5cAxQkcL5t79ImLFpVl2E1ijPQ/s+Zj4XJ4PWVsieP0n95j2Sxw8TFVfixMOFczRu3iJx9BRWNkP9Stzzmr/4Fs1bN0EpvNkZBr/1IvWPP+kcU/LCecLZR4SVEH3oMDSaZN44h//gzromoOsiL5wheHCv5/dNnT+PlZC41bXOa/bkFDKsY+YLhNVYaso+cgJdml93dQdHkVFr1oNhIBwLXSkjfA9ZmkENbC5delFaeBvd32vXrqGU4saNG4RhSD6f7wzv6bb8XmIL788Bo8Df2XCv/o0nubcvJeF1N/6/9957Ow7w7kdrrt0idu7cuc4T8WnQWm9JaBDHuzbGz9rdCXEMT2zyOk0JNV+hDEG4hStpG9AIIhpabEpoQOzS2lLgBorqVr4oYLWIrNb0wTTZaI8KNJYQhJGiueEnisksVm3xol531mxZgLVA0Qwh+/V/g6yUsC2Jn3QI1qoIrbCGBsH3iCoVEpOT1FcapE8dxV9cREhB6sQxmrfvYA4UcApJVj6+ijAM8u+9TuN67N6agwUSY0PQCuMK2yZ74Szu7VsdFzaTTeEdmyB8eLdz/ciBARLDeTzld1x+YZqkL5wneHQbZZqIdAZdr5E4fRbKCxCFiNFY9MI6ehKqy6A15pFTREtzWCkLVudhdAIjnYXSPDo7jHSryNXHqMwYbGg7exnEP4UQSCmZnJzsuLflcplisciDBw867m+pVNpTHd6XvvQlfuiHfogoirh3794XtdY/tmH/DvCPgPeAVeD7tNYPd7MPrfXfAP7Gbj7zQlxawzCe6NLutjd1J/vbDlprpqenmZ+f31WLmO6K13VThtKaZouMut+TQD2IuysM08IU6+UephQ0AkW95YeGG8pKLANqXki5lRFImr03SztmVnLDOIOLQsr1J7QAEqagGSqKbkyESdErDOBIQaAUZS8+xpGU0TkGsxUfrPmaeotIk6aB0BrLEPiRotI6dq016eYa1vWvdB4+ZsZBNVxkOk1UqyIDH2dyAm9mhuTRk5S/cRWZTOKMDOI+eIg1eQgZ+Xi3b5A6dQwzaXfILnnqBLpexpueJnHiJPbYGFYmgXv7Vuf7pi+8Rrg4gyks2leZc+Y0enWRaHEeQwiidBpp2ySH852yE6II58QRpGmgHq9bgMHMQ+xT52F1Lm6vA6JyEWvyMKLSknHTEiOsIZRC2wlE6KFCH6s0TTBysiem+jIQ3sbjMAyjk92Fdff3J37iJ/ja177GX/krf4U//If/MN/1Xd/11Nkz7QE+X/7yl5mamsJxnO8XQvyi1vp612Z/GihprU8JIf448OPA9v15B4AXctZN09xk4bXVVW7fvs277777XLJC7all1WqVS5cu7ZrsNL1kF0TrZAcxSQhiAqsFva1kQrRidEpT9qIeSzFQ8XuWhEYQstIIcLvMRTeMS1cSpqDeaLDSDFlthutFxci4CDgK8Bs1Kq7PUiOk6ndZfSpEaoUpNM1AsdwMWfPWj1EpjSMBrVnzIopu1EmWOIYgVIpmpFjzIhph3GebMuPvnPudfwa+H+dOEJgpB5lJE62VkSrCGh3Fm5klcfwourqKkc9h5dJ4s3MkTp+CRoWovIZ96BDpo+O4D+JkRPbdNwkX54gq8cAdI+lgWhAszMfnO5Eg+9YFwpm74HskjxyOrb+33kDPP0IEcYGx0JrE2TOYVkS0srD+mySSWJkkev5+z8PTOXkGM53okB2ANTKKaa/bCzKZBDu+foQ00FYiloxx6xjN3u6fl4XwtktatN3fn/mZn+HcuXP8xb/4FymVSu0pZNuie4BPKxTVHuDTjT8I/HTr7/8U+P3iOSiivhQubbdA6EE2/m+HdovYxuLlJ6EtgtiO2eneN7d0YZWOXdSNP2Po+yjTpBFuTqa2M6luqKgHm91+Q8QWoRdq1rwAjA2xTa1JWZIwVJQjCeYGEtcKGXrUfB+V2FzaYxsCA2iGCq/bXdeatBW70iU3ImlYKB2TsiWh4mtcT5O5/zHJxRloTS0ThoGpNVGpjLQkVqGA//gxiZPHCWYeYY6MYOUl/uwcqTdex79/G5QidfYswcIsMpdFODbZ187QvBV3VQjLInPhPN6dmyQOn6BWKuEcnsIwFP6DO+vnUmiSU2P4D253CEzYNurQBImohhusVwqI0TEsM0JN30YNDCNLKyAlidPn0EszqEy+s61z6hyy9LiTlZWjkxi1JRieREuJDBqozGBsTYQeZn2VyEpDK177KhBeN+r1OpcuXeLbvu3bdrTuDgf4TAIzAFrrUAhRBoaAlR3tZI94YS5tm/Dajf9byaE/K7Qzv6+//jr5fP6p23fHBrdKUHhPILtGoJAq7OmvNARUlUSEsZvpdX3QloK1ZoivYtLqRpvoluoBkYbB5IafLgqxJKwFUHRDhjb0h1kyXn+1ofCVia0D2ltopUgYEGhBsRW4KzgGiHVCK3ua5Vblclsy3jEEFX/9tbRUjH3yK3HxtI6lpHQYQrOBPTKAhpjsThwnmH6ANTkJ9QpGrkD67TfxbsUeT+atN3Fbf9deg9SRQzRbLqs9NoqVSeLduRmfF9sk89Yb+I/uEqn1B0Tq9deJ5h8Qh4pa53fiEKYj0OVFIsAcGiFcXSZ1/gJ6eRrhxd/FzOaJGjX04CB6KZ7LoGtljIkjmOkUshRncfXKPHJ0CosgvtZX52HqJNKrosvLMHIYUVuFZAbTqxIaFrRKp14WwtvJcXie90ooOO8EL8TCa9fLzczMMDs7y1tvvXWgWaAnSVJrrbl3796uW8Taxyuk3ER27ZKTdrxOAF6oO+6f53k4KRNDxK/XIr1pXqstBVUvZK3LHW4ECkuClLGk+lIj6Nl3u/eW0MdzPRpGosfq9CMFSFJm7H6uNHpjptqwMFBYUrAWKOrhesRR67jjw7EM1rxofW6F1mRtiRtqGqEi0KJl+UlqgWLw5q9jhm5cXSw0OogQgY/IZDASNu6NWyROniB4dB/72HHU6gIoRWJqhOrvfgVhW6ROn+qQXfLMGVRxATOVw6cVm5ufIVhYA+KaOStj4X50DdHqJjIKAyTGhwlnYkvPOX6CsLRG+vXXiebvo911t96eGMcZHUIvPeypjzRViD02iq6uu6JaCGpCMljqKlmxbMyBgbjuDsA04+ytBxgGmDZEHugIw6+hrCTKSe+r0+KgsRPCayc4doodDvCZAw4Ds0IIE8gTJy+eKV7IYyaKIlzXZW1tjUuXLh0o2T2p+DgIAj788EOUUrz//vs7JrvuNWNxzHXEDfoxDLGemOguDLaTaSRQ9lRPHA7iYuQoUizVg57YH6zH8JbrAQv1YBPRCsD3XIqBoL6B7AStvtsoZKHmbyK7tCUJw4Cyp1hxFaFudcZqTUoqdOjzuOZRapOd1mQsgRSw1Aip+IpQKdJmnLFdboao0hJjqw+RhhkTnlKIKELkcgjLQkpwTh6Pye7UadTSHMK0sCanoLyEMTBI4tA43r07ICSZt94gevwA7TZxhgtk336T4OFdtBfH4ZJnTmOnTYIHd0gcO9x67SyWownnp7vOhSZ97hRq7m7PnFz78DFMU3UsuM7rJ88hvSpGrrC+RiJF8thxBv0SuiXTpaRBlB+A4hzaiEMDYmQKWV2O/50bQTbW0E4aI2jG/cmhi1DRS2Ph7QR7mVjWPcCn1WDwx4Ff3LDZLwI/0Pr7HwV+TR/UOL9t8Nxd2nbjv2EYvPHGGwe+z7a73B2baLeInTx5sjNXY7draiF6sq5eq6G/DaXjkoxuYorLUDa7wHFTvWK5rkjbvRd+O6O6WPfJ2samIua0JfHCiMdVjyRhbEm0IAXkbIPlus+MqxhK9P68OVvSDBWPaz5gMWiBr2NSyFiSiq9YchUQu7OO1BCFVAPNUmi0jl2Rtw0CBcvNsHVMgnMPfxMdBGAYcbJGgcjlETqCSglzeArv5k2cs+eIHt1FDgwjbQO1MIs8dAQrWSdYmMfIZklMjuHfi11Wa3QMK+1Qb3dZ2DaZ82fwH3RJNw0OYKTTBNN3e665xJnzUF4k8roI37KIxg9BdQlVExjDY0Qri4hEEmfqCHplrvU7xOQoB4axUzaiEoeWjIFR1PIciYlDyMYaaGgkC2itKDRLCCGIEhkMIVo90a3fIIqQoYcRxYT9qhBeG7uxSLsH+LRCV/9kiwE+fx/434UQd4EiMSk+czxXl7a78f/q1avPZB8bFVO6W8T2qviQzeUQXWUeodI9vbBxyYaCKAAZP+0NEQf+QxW7rG1YUrBabRDJzQmDpCkoNQOKbrx4LYg6AncpSxJEivnqupRROpVizYswBGRtg8Wax1pzvYLYNAQi0uQcSdmLmKn2tvOZRDi2xWozouqvx79MARlLEmmohBJaKitZM84oLzQ0KUJySRtDSpL3L5OoFWP32zJRQiJyabQfIOoVxMRhxMoK9tnzRI/uYExMQaOCLtUxTp5DLDzCyGQxshmEXyeYfRSfj3OvEc4/IpipIkwTe3wsllfvIjvn+CnMhKR2d72MxMgXYpd0KV7HnjyBe/sG1sQU0tRQW+lki4zCYCzbLhR6Zd1djRbmsI6exGiWEG6t87quV7DGJ5GNdXc3ZZuYho7JHWg06mRlXMsTKIUh7FhUVYAMmhSy6fWQxEuOvVqjXQN8AP46sHGAjwv8sYM4xt3guRDeVo3/z2pyWVsxRSnFzZs38X1/Vy1iWx37+KH1uqN2k3zb2uuQHTERShmTWqUr9uWrVn+phqW63yFFaEk6SUEjUjyu9WZlQwUDCYNmoFjoIro2/Ehhhk3KkUmp2dsqIUV8sH6kmK70urSOIRCBS8O3qHldBNkiuqVGwJobcnIwidaagmNQ9SOWXA1IbAlJ06IeCUTT5a3HH8WxESljlzaVQnseslmDiSmYn8U8fBz36x9injiNnp8GrTBOv4aYiedJJI5O0LzycaymkkiQPH6cYLo1a0IIchffxb1xpZOYEIkkqVMnCafvEQIynUHVayTOnIO1RfTSbOd7CaFJnH8dtfCg9wRKiZFKwMocPQVGUmIfP410LHRteX2ddA4rm0I6Np2G3GQ2NrLTeagV0aZNNp1Bp/LgVpBaUWm4SG3S1qx2DJtIv9gpZjv1HhuNxsvaZbEnPHOXtt34PzY21tOu9awmlxmGQbPZ5OrVq4yOjnL+/Pn9BYiFIApDaGWpmq0nc1vost5V22Zadite10tcpojl2ksbOu2lAFMIGlFExet94ltSYEqoehFrbi9hGQIyjsFMxcNG0C1rJwUUEibzFY/ZIIr7V1twDEHKFMxWPEIlGXIUSLmJ6KCV+NExUc/XY1KUaPK2wXzNpy4Eoxmbd+Z+B1PHcySEYSBME9VsIv0memwSufgYOXkUsTCNdf4N1J2rkEwjR8YQM3cRtoN1+CiR54OKsKYOI5VPMB037xsDQ1iFAtrQ0CI75/gJpN8gbG0DkDxxAuV76MVedSBzbBLTFPhLMz3xV2NkHCuVgKVpzKERotW4gFjmB7HyWcTaPKIwsl44PjSOZSiEW4VUy1NIZDDSaWToro/gLYwj/Traq6MSORxDYKOJEg6mgDXXx3AMPNdjeXmZgYGBPT+M94PdqB1/WiaWwTMmvCiK+Pjjjzl37tymxv9nNbksCAKuX7/OhQsXdtwith0ipXESiVax8bodEJedRHTrnxiCTS1fthSsNDYrw9hCU/EVjVCRT/R2RWRsyeOqT9gqT2lnnSWQSxjMVX2WWxZdRka4xHGzgYTJfNXnXqPZWW8gKZEiJrqZssdS1+G5kWY40Ut0AFlL4oWKqhdSCeOg9WDCYLURMlPx0VqTT5pk60ukS48xDSM+C0IgDYEMPBidRKzMI0YnYPkxHDmFWFhGjIwjtEIuzWIMjyFNSTT3EJ0pkHrtdYKZu52QhHPmPGppjmhpDmPiCMJJkDp9inD63vo4DiGwT5xF2BbejfXpY8J2cI6dRC3OoD2wJg4TzDwAaRCOT+K4FajG50nmB4hWlzCPnsJoFBH12F3Va8uIzAAyX8B0S4h20La0COkBjFQCGcRriMoyeuQIhl+P/+030bkRiFxAIC0brRW5pEMtVAzls8wsrvDo0SOklAwODnZUTJ5HBvfTJP65GzxTwjMMg8997nNb/oDbtZftBVprHj16RLlc3lU/7NPW1LTkrDyPoMsVdf2ARr1OMhOLkBoizsQCnbYxS8JifZ3sMrakESgCt8GCWl+r4kZYBiQtg7VmyHR53c1sBIqUiEgnHJYaAStrva5rpCFvw4qruFts9rznGIKMKZip+D1EJwWkCVis6o5IJ8QqK1LDbDl2nxOWJGsZuEHETDn+HhlLopTmUbHBd7gfYwkFIh7saIQ+kTDQw+PI4hLG4DBUishDR5BL04jCIWRxCRH6iCMnobiACgNkJo85PEzt+nVQCpnNY4+NEc2tu6CGY5MYHeix6oyhUYxMGr08A5l1MVjryAlk0EAvzazP90gkMUYmMByTRKPc+zu7dZzjpxGVxd4LwDAwDx1GrjzsvYZTWcTwGLK83qWhM4MIJwVhLCQQFx63ahSTWYzQJbJSSHQsrGAZnDpxAnXiBL7vs7q62lExyWaznTav3VQT7AZ9wntGkFI+cXLZQYl1hmHItWvXME2TycnJAxtz151djVjP0vquS2TYpDLZzhyJtS6X1JSiVW7SS+gCaPgRDdVr1VqGwDFiC2wjklJhmhb319xN7w0lTRaLLotubxbNlIIBW/Kg5OJI0ZNNHkmZLJQ9HnX6cgVeBMnW/tubpkxBEEYdN9wQcTHy9Fq8zXfKBzhehbjaTyGUii03J42af4yRzYLnIgeHkKuPCSdPIGpxeYY/eYxkqxwkHD1EImigl+cwh0Ywsln02jJRq7REJNM4h4+i5h8hnWRMIaaJc/wM0dI0uhRbVLpWwZo8ipFMopZne1tYLBsjlYTVx4hG77VoTB7DCGoggt7SnvwQViaJdNfQ3WsNjGOIEBGsP1x0egBp2wi3Glv8holhmsigSeRkEJaDIKKnqKmjHh2PNO1WMalWqxSLRa5evYpSqkN+uVzuwEJAOw0n1ev1vkt7EDgodeKN8y3u379/YCKg7VkUYRiCNDEANwgIhNkiP4ElodRVzCoAN4xY2xCvS1mCR2W3J2MLkHMM5qseGbuXpKWKsA3BbC1kcEN3WN4xaHgRNxfj2aijaUEjivc9mraYLjZZqsTHVPdCQDCcNCk2Am4tNTrrGCKWolpoBCy3FZTRDCZNHpY8Fqo+E4UEw0mTlXrAo2aIRHPcdDnhLkAUYQgdt6uh0bkBqHnxABzTwjBE3Bc7fhR7eZowO44YHSe5+hhMC+f4KazHD+Nzbdp4iRyJ+Ued47OPn0HU11Ct18zBITBtBBFq8dE6fQiBdew0QkrCrjYyAHPqOEbYgJVpGBhCleLyEpHOYg2PdJr/xcAoUTmWgDKmTmK6xTheB5AbhsoKYuwo0i3HCaFGGZ3OgwbpOEitIFKo9CDCdpCqZdkLidStHkIhUEgQGklLRHWDJKwQglwuRy6X49ixY4RhSLFYZGFhgdu3b5NMJjvu73Yadk/DTodwv6wTy/aKZ054z2qQD8DS0hJ37tzpaRE7yLkWHYvODxCWQ6gUQRghWrLqsRxSr2KK1nFyImPLTvbWNgTTLestZRv4bkwUlpQ8KseWW8kNcQxBoDRm0KQUWZ0C5mIzJOsYSCGwBNxdbtANW2gyKZulqseNxXrPe0prHKG5veEzg5ZiualYqQexEILWjKQslmo+d1fjY8rYkgHH4NFafOwDCYO6r7jo3sAyDaLWg8UAyOSRYYAwDHRuEKO2ik6kEckkVnGOcPQwZhARlZYIswVS2TSqRXbm1HHMZhnDSuEDKplG5wdgZa5zXkUqjVEoEM496LHejLFJTEtC6TFiZL0nWuYHMQcKiPLS+ra5Aqq0gj98iIxuriudALpZg2QGa3gYo9HbzikSKcgcx2yUeo00Jxsr0+guS9604lq7Tu9uIq6VVCFIiZYSgWrNEFYgticd0zQZHR1ldHQ0HgPQaFAsFrl58yZBEOx5glnfpX3eO97HIJ/txEEPcuZt+/mrhKQtlmTZToeIQhWrpsTSS7F2StsFDKJ4HkSkFY+r69+zGShMHVDxoTsxqwEr8qh5ilXVq1NntAqKby83Ng0GyhiKhgf313qJLm1Jkpbk1nKDE10m4lDSoOlH3K/Ex5lwI3IJE6U1d1eb7S/PZM7mftFlPOdgCM1wyuJByeNzcpZxOwIl0DpECoFO5ZAozKABzjBm+R5q+BBGo4yoNwgPnSRRnME3HMxT5zAXZtDlYqxOMjGFarm30krhnHoNXVqAduIAgTc8TtKvEpUW10UAUlns8Qn06hx4rU6RyipYFtbhE3GWtYvsAIQUGIemyDbXNv3eMp3FTCeR9V5lE9J5pGMjqr0kqHMjGAaIcP1aU5lBDB2g7RQiaKKSOQyhUEYCJQRSK3Tko4TsXFtx9HNnSQohBOl0mnQ6zeHDhzdJuFuWxdDQEIODg6RSqW2THzslvEaj0Se8g8Beicn3fa5cuUI2m91SHLQ9n/ag4Af+upUhJUGgkMTVHm19ulDF7kmxK9OpdDxIZ7XZS+qmFJT9CE/3ZmYNv8FyJHA3/CSDSZOlisdSK2vbRsKUZG2Dm0v1TpmKQmBKwUja4vZyvTOg25KCtC1JWwa3V3pdWhPFfMXrJC4GEwZupLi1EpOfLUEKwYOSx4B0+fecVbQWcbxOA8kUUkrMxhpufhyrXsOfOI61OoNKZNCFQRLFGcJEmuTAEPWr1xBozKlj0Kysk11hGDObw725LplmjE4gTQOr3CKbZhXPThBkB8ioBhQf98rQD4wgC0Oturou2AnM8bjtiw0KNCI7gJnLYbhrqMIhqK+tvzl2FCOoIRslVHYIqnGrpx4+jAw9ZOij0gOIegmVKmAYcSY9Mix06CPtRKwsHbiETqYzhrP9MwopEHrzSICdYqOEu+u6rK6ucv/+fZrNJrlcrhP/21j6stMYXq1W69fh7QY7UUzZKSqVCleuXOHUqVNPbBE7qEE+sbsmCIIQdOyWt3thTdmbpJBC4HcpdZhSUPNDzK54ndGquZsuu+SM9Us8aQiKlRpVFf8UeUfSjBSOIUgYklstF7XihiSseGj2eNbm7nKDuVZNYKRhIGliGwazZZerC17PfkGzXPOZi9bFASbSFgs1n5srPqeH09T8iOGUyb2iG2emBYxnLCpuRNmN3/u/RjMYKsIyBG6gMRJJMC3M+ipBfpRUfYVKahR79jLh8CSWV8WortAYnCQbVBGlReTAEK5pYZZaGU7Twpo6Biuz6NUq2AmEITEnptCL0z1kYEwcJiOMjivchp/KYaeSGPUV9NAUnV9CyDgp4ZY7GVhdGEasPI4TC4eOYjRWke5avHnUStok0sjCMIZXWd+JZaOFRIwciS3Z9k+rNTqZw7BMZOtopVdDZwfj+CbE7q20QAcowyZv2ygt44lwHVtv/0gkEkxOTjI5OYlSikql0plh0S59GRwcJJvN7jiGV6/Xn5uK0fPAC3Vpd2OJzc3NMT09zdtvv73tE+cgXFrXdbl3/wGnz54lmUzSDBSe66LMuPg40jGpBa2RiCuNALv1+DYFNPywowzsGPEM2LofstIqSK5FEik0aRMelJoosf4z5BIGGW3wYLVBs0umPdIwkXV4XPa4Or/e6gSxW7tWWmMp6pXwOZSN43ofzlZIOyYIQUJEpByLW12WXs6RVP2Iu8U4djecMqm6ATeXG5wfSTGethh1Fxg2K1iGpGXjguVgNkpEmQEMr06QLmB7DfxDJ0gUZ1GGTXP0KLlKTG5idAqzVCa1HLdwGROHkaELy7GVJ6TEPnEGtTjdU0AsB4Yxs1koLaKHpzo1eCKdwxwcxirNI1r9so1aBQdQA6PYjonR1SkBIAwDPztEJmMjN8TqqK7GVl3YQHaTHSB8F4YPYwS9sVCBRiTS60kKQKULSCvZqsGDyMkgiTtvtLTi6W1tZZoDIruNkFJSKBQoFAqcaJW+FItFZmdnqVarSCnJZDL4vr9t6UvfpT0g7JSYulvELl68+NSq9P0SXlti/sKFC50ZFBsTL41AY0lBgMZtCQa4oSZjS0rNkFoXUWVtg0dlt+NeQut57tW5V7NArLsVSVPS8ELuF3tLUGxDUHBMFsoui7X1OjxLCsYyFtcWqgym1snOEXGc8Mr8+gNlLNnO+ioW625nf4Mpi5V6QNmNM7CTeYcbS3V0y2oMIsX8WoU/XljCUhIlTUwp0a6P4dcRTgItJKYhIPLxk3kSC9fw82OYhGQrC5BIIgdGEKUFhJEmtBIkJw7B6nrvqhw5FI9/FALfj49PpNJYY5OwMgulmGiEimJZpokjiLUFxNpCTxIjnXCIDh0hUS8i3A1kks7HqshBGeFvkMtK5RGZAtJ2kKVestMDE/F+bQfC9XIUlRvBQKGdJDRjwlOpQjwBTgWgNZGVjDO4CCLDXvd4pGSH3V0HAtu2GR8fZ3x8HK01d+7cwff9TunLwMAAg4OD5PP5Hlf3JR7gsye8MMmGnRCT67p88MEHpFIp3nrrrR214OzHpZ2ZmeHWrVu8++67naxv1FpKC4ltxDEy1eqntQU95BZEuqcJP2MbPK71xt5SlqTacCn31g8zkrJYrnrcWGqQd9ZdjfGMTdUNubpYY67i0faSJ7I2QRRxeb5KpGMZqZwtOVpwKDYjpruMwLwMCHyPB5V1KfkjhQRuEHFzqU6pGTCUNElakuuLdZTSHB9MsloPuLHU4M8ebuBoHwwLU0qilqCpQUCQzJHwK/iJLNIwsIIGzdFjJL0ytlcjHJ7CsExEaQEQmIUCdsJYJ7tMHnPyOEajBPU1pGmAYWIdO4PlmDHZtSElMp3BymaQpccdDTwg7mmdOoElXGxT9IRSlJWgmR/FNCKSbgnfWr+BtWHB6HGkZWJ4tVapSOs904HhIxiRj9RR5+GkEeiBCUyiWBkmaBUbJ3PI1nhLqQIiKxGv3zoWLeQ6OUtr04iA5wUhBKZpMj4+zrvvvsvbb79NLpdjaWmJr3/961y+fJnZ2VlmZ2f3VZZSLBb5ju/4DoQQd4QQX26NVdx4LG8LIf6dEOKaEOKyEOKZzrV4aWN4xWKRGzducP78+c5gkZ1gLxZee1RdFEVcvHixK7YRz2sNI+IaKr0uvCkAL1q/4ZKm4HHNJ9MSw4zbw1wiDcMpi6V6QN6WPCo1CDEAg0LCpOZH5GyDawvrDJVxTPwobiu7PL8+btOLNEcLNs1QcW2hdwzneMYmYUg+nF23ThxDkBU+9ypwYiAudrZFRELozv4EmpQhmK+4eFGsYjyecbi+EM8t/Y9OwmBQQRrx4KFIGghhYLpV/OwIqdoy1ewh8s1lmk4W306RXbmBZ6dRmRyZWitTWhhBSpBKYbSttPEjiNI8rK13LEgngT04AKtdRCcExqFjyKCOKD9GWSa0HxhO3EEhG6vI+nL8w6RzqEoRDBMxOoXtlkmoZodsfNPBCRvUEwVSCbs3Vldfi0kqMxCH3vz17LdoltGmjcgOYXRNPpOhR5gewDQk3WWWynQwulxXIWMlmdD3sR3ZIryDi+HtBt29tKZpMjIywsjICFprms0mxWKRH/qhH+L69esEQUCxWOTbvu3bdmXt/diP/Ri///f/fr785S+fFkJ8Efgi8F9s2KwB/Cmt9R0hxCHgG0KIX9Zarx3MN+3FC7PwnlSWorXm4cOH3Llzh/fee29XZAeb5aGeBs/z+OCDD0in07zxxhs9gVxBy52NIrQ04hGLrYvcMmLrLmnFXRLztdilsUxJ2pLM1/xOh4MXabIm3Cs1CbtOedY2CFpWVjcipVmpedzY8PrRvEPFDbi9vP66JQXHCgluLNZ6pKEmMybNpse91r08Xws5krNxI8lcKwyVMRRpHfDJXJWsJRjP2BhCcHO5QcIUvDMg+aahCEOFGEKgEAjTxopchGViulWa6WHyzWXWshMkREQirFMZmMI2NZnGKtpJIsaPxImDRhmEppEbwcpkEMW5eNANsXimMTSK4VfAXf9+xqFjWKNjGNUlROt1mRuMuyemTmBlHMz6csttbP1uQiPGjmDmC1iN1Vhmvwu2bcPoUXKOgal7H45amujRY0gVItWGB2cigxgYj8muCypVwEikesgudHJxMXbr946suExEAE2/e5jJ8yc7eHJZihCCVCrF1NQUP/dzP8frr7/OF77wBX7zN3+Tv/W3/tau9vELv/AL/MAPtDU++WngD23cRmt9W2t9p/X3x8ASMLK7b7NzvFQxvDAMuXr1KrZt73mYz24Kj8vlMlevXt2291aIONBNECANaERxS1Cj5bpKAWVvvS3JDxVVL+pxY1Xos1BuAOvB4ZyhuL1UpxH0ZndH0hYfzVWYyjvUW0Nmso6BLeHDuQopK7YiFILJnMNixeMbLatuutRkOGOTEBHXl5q0b6acYzCUiDst6n6E1prTQ0luL9Zxwzg+6Sif+0UfhWA4IWn6iu89GpE0dOvG12AnESpEmBYhElsaiMinlhljwC/RMNM0rQQDpVtxO9boYYx6MY6zAbowjmHa5Fcf006lisFxpCEQ9a4EgmkhhycwCBEbkg6YNkZ+ECOsxRZdF7QQiKFJpBQYxQU2QiezBIksyciDoLnhs5IwO4xFSN1tku3yTLSQUBhDRh6665rUgMoOYwoFkd+x1UIn2yG/yDDjCXJta1ArdGvwUnyFvFyEtxGu6/Kd3/md/Ik/8Sd2vY/FxUUmJiba/1wAtlXfFUJcIr5J7m233X7w0ri07RaxI0eOPHXu5XbYqUvbzvq+8847T+wVlCLOjgohiAIP0zLRgGOst5MJwJCSuNsW3EjhmAJaRkBCBzwoK8bSaZotmaWCCbeL8d+PFBJMr7kMJk2qbthxNRNmfDEeLTjcXKzRaMUKG4HitdE0Sms+mut1aw8XEtRrTW4318n2zHCKB0t1bq65vH04T9pWDDgGl1ufzdoGQ0mLQBgofI7nLW4vN/lz5wRTGUlQrWGhEZZNKCRpEVIiSSGoUEqNklMN7KBKMTnKQLCGiBT19BA5R0K7FCQbN9Yb9RKRap3r/DAykUBuLOgdPISVU4hulxbASiBHJxGNMlqE6GjdmtVCxkSnAqQXnz+VyscWJaCdFCI/inCrJJQHAqJUAdGuq8uNIE2DRBQAgqwh0FGsWNyQDmYiRbLlwmqvFmdWpURnhrBEq0dWKyIrGbeStdrIIE6yKDPRITwlLQwzvj7VCyI72F2nRTabfeL73/7t387CwuaHy1//63+9599aay2EeGLIUggxAfzvwA9orQ+mN3QLvBQWXrtF7I033iCXyz3lkztfdyu0xUg9z3tq1ldKiYoUaI3RFZ/ROs5wNkNFxVufa5G2JDMVD0Ec0wvdJtPN+EOL9YCcYyA0PcW/XhhxOO9wbaHWYxUWGwEjKaMnJgcxQXp+xNUudzdtG4wkTS5Plzk9ZHVem0hbXJlZVwYRWuN5Ibdb7WzHBxIslj3uLNUZzVqcGE5zbb7Gt4xpPj9hIqTECD2kFASGjaM8mkYSy/MpZ8YZDNao2nlMoRkKSjRlgrJMMRotxpGZZAaRG0ZWliGMM8a+hjA/QjqoQhCTkwYYmkIKjWxWwcmuB/OdNHJ4AlEvrXc7uLU4/tUmushHer3kTzqHDjwYGEd4tfj9Ln4Rpo1O5hCpHGbYXM9OAUKFqMwAWA7pyOuQVfscliKDTDqHIza7w6KL7DRxwiSuwWtJiZk2WTseBvUiNY93Wofn+/62E8t+5Vd+5YnvjY2NMT8/z8TERJvQlrbaTgiRA/4V8F9qrb/y1IPaB15YDK+dTb1z5w7T09NcvHhx32QHT+7dhfjH+8Y3voHjOLvI+gosEU/xgnXhT0PEsbuqH9EIFIOJWJATWhd6o8pcc/1GsQ2BLQR3NnQ6aKWZK7s9ZDeZc1ioNPG6MsCmFJwcSHBltsKtpXqnyPXkYBKvGXD9cXzDrzbh5GAC5Udca71mGYLXRtN8eL9E0w8RaM6PpLmzUKfcDBlKmUgF82WPiZTmz75mIYREaIVNiDIdhBAYaCJp4UuTVFhn2RmOLbzIY8kZwTYlw8IFy0GMn4hjVpXY7dSpPI3UEAkD0m2iEwI9cgQxOIHhVhDNFmkZFjqZQ06ewkjYyOoyojsOpyIYO4EsjGB4VWTYW8+prQQkc4h0Jn5/Qy40NBywk3FmNux1bYFYACBTwFR+D9kBROlBciNjOGbvrdM0kvG119UbG1kppJC0nVdl2p2JdS/SnYWdd1rsR5vve77ne/jpn/7p9j9/APiFLda3gZ8H/pHW+p/ueWc7xDMnvCedMN/3aTbji+299947MN2vJ+2vUqnwwQcfcOzYMU6cOLHjH1IAhhG3UsXu7HpZSjeaXTMKjMhjVTkkWzeFJQUq0txdbdAeN+tIjSPh2mKNXKsMRQBHCwk+ni2z1gw7JDietXEEfDgTW3t1P+LUcIpTAwmuzJSptFrabEMylrEpld1OAuNQ3iFvGXz8qEwQaU4NppjMJLg8W0EDJ4eSVBsB08UmU1mbL77tkDAllmUSBCHSkATSIqma1O0cBVx8JcG0GI3KrNiDBHaaMR2TlZEfRiRSiPJifM7SBfTQFFKFpJUbE0giDWPHkANjGI01pLdurapUHp3MYFoyJrou70bbSdTgFCKVR9j2ZqJLZNBDUwgngRG6yA0PPu2kcNPDmMkEpl9FWL1qIyqRRedHMYXCCDckJkwHnR/FsgxMFcRxPeLsq2tlcOy4xq7ajK1nXzrI1hwUqRWRMJGt6WZa6xfqzsLOXNq9TCzrxhe/+EW+/OUvI4S4A3w78GMAQoj3hRA/1drse4FvBX5QCPFx68/be97pU/BCXNp2i5hlWZw+ffqZ7689yOdpXRpbQYi439E2DQTxUJZ2xajSsZWWMCXzVY+CqSkHioYyCLViPGMzX/HQSvO4Zf0dG0iw1oiTGNVWGGqmFcMLI8UH02udfT8sNrkwluHrj8odqap4jSTaj7jc1XFxuJCg3gj45FGZt4/kma8FXBjPcHU6JjqAUyMpIj/iwWoDrTWvj2f4eLqMBs6MpPiDR2AsBbZlo7SAMECJuGVKmw6ODlgzMlhUcQ2HppFlRDeIEMwZA4yaAaJZQYQ+QSKLkcphNNbWhQBa4wuFvYaoFTvHrgEK47EL3SgjI6/9anyeU3lwMkivinRj0hfGuqagzgwgnBTSrSFaMTwCF2UnEL6LdtKQziP8OikRy2VBK7YXemgrgU7l4zGKUUx0IvJRVhIRNNGZIaQRd8fQ+nRoJpCRj3LSdD+qs8kEjUiTsNePL9IaZVoda3FpZZWB4dFtr7vngR0/9Pdo5Q0NDfGrv/qrAD03udb668Cfaf39Z4Cf2dMO9oDnTnjdLWKffPLJM92X1prbt2/TaDT2NchHSEGzUWcgncGLBFqC8OOylMGk2bGw3FDhGAYrrar7UjPAFPCosm6JKBVnU7vHXgwkLbSKuNnl7mZsg7wtqTWDDtlJAedG0nz0cA3LiC1PpQUXxjN88qhM1LIIK42AE4UkHz1Yi48fzWvjWT56tMZEIUHKloynbT6aLqO15s1DWU7mNOcLIbZtEel4X4mgimemyekGZSNHQTcRIqKiTUZkiNIhC8YAWSNiSrTaqFI5IiR2swyNeP++tJGZAUy/hqgXOxPbtGFBYQwRuOvacwB+M4535UcRhkQ2a4gNMTqtwjjZYJgYQRO8ek/HBYBOD0FGI/w6Mmhseh8hUPlRZOj3SDp1Pm+nEKks5haxdi1NtGltcpG06WDbZsdbVVoTCBNTyLgMSwhmHi+8FIT3NOw0sfEq4blladvFvWEYdpIFz2pyGcQu8+XLlykUCrz99tv7ikUYUrI4/5iR0bG4/atVWLzmhTQDTaneBGGSSjg0u+SCkoakSxuUiazN1YUqExmbh+W2xZfkw5kyoxm7M7viSCHBQqnJ7VLAaDZ+fThjY2v48OFa/P0ieOdQhnIj5MMWsQGcHEkxv9ogm45JZSBpkrYMPnoUbyOBAdvg5kINS8Kp4QwycPkTbwxgm3Fnp5QGgeuSNCwkGs9IkselaGQYEB6Odlk08qQMmOgQXR5pWLGV1YwTJQ3pYGfy2G4F4a4nT3wMrKEppFvtzI9oQ1sOIjcMTiqO6QX0EJE2bcgMYqAQtRKoDS0rECuY2E58PNXlTUQWSRORHUJKEF5zEw9qw4otPsNAblhfCxn3xrbaxbp7YiM7jtlpYbSSFDFpOq3OlDDSLK+WqFarfO1rX6NQKDA0NLRrLbvnhU9bWxk8JwvPdV0++eQTxsbGOHr06DOfXBZFER988AGnT59mdHT/T9K4v9PHkoJQaixDUPU1SUPQcJukbJNKAL5SVLwollhPmFxfqGEbAseUDCZNbi3VCZXGa1liRwsJvtYioscVj2MDCVKWwcfTax0JoaWqz7tTOT6ZqdDsals7O5qmUQ+528rWSgGvTWT48H4JDRwfTZN1LGZXGyy25OHPj2e4MVPhzWMFCkmLjCVZLTf4yT90hIxoIIgnpTUigYoiGjLBoHRZ1mkMI2REuCypBNp0GBcNlIZZlWYgYZIJPQg9FOAlCigVkiaElguqEejccCuh08DYSHS5YYSdQDYrCLdCZPbGdHVmAGEl4oxrEH/ntssKscWl0gMYUmOqEEK3Z5obxDFAVzqkbIkkBAWRnUT4zdYaBipdwEBjCAVKoYQRt5URJyEwjDgJIgSRtDCUjxIGykp2XFbRIjtlJuKxle3rSBrYiVix+OzZs6ytrXXknNpadkNDQy+NpPqnTRoKngPhKaX45JNPOH369KauiWcxuWxhYQHXdbl06VKnH/agIKXAMUU8ZxZF5NZJ2AlKjYiRlMGtYuySHsrafNKqc/MjzbFCgq9Nr8fhVhshQ9Lng+l1V9cxJTnb4GstCw7izOzpoRRrVa9DdqaEMyNxxtUQUMg6mIYka0m+cT8mEUGsQnx9LiYb2XJp25ZgFEZEYcRK0+N//eOnycgmWomOjoGQEjNsog1JTaQYlT6LYQLPEow5EeVqwAw58mbEYSMCInSrzEN5DZxGZV09yU5BpoDwm7FbCfEN7a6hLQc/kcNQPnbkQXP9fAjLiV3e7BBChxihB36tN7HppGNrKpVDRC6WDnpqPUTgxq6n5SCSWaTyyWx0W60EOnDRqQGkAdaGjK42bVTko+z0pmyvQKGkhTad3tIVIDQSGOb6da10XKnZ9mi65ZqATjvX3bt3cV33mVp/O01EfNrUjuE5DfH5pm/6pq13vg/V441oqyBXKhVyuRzJZPLpH9olhBAYQqK8KpEXkM9liHmoNRya2MGpeVFsDSpNPmFye6UeS3pHMYllLMFaOaKdJB9ImohI8cHDNfIJk7IbMpSysDRxvE4KHBNySRtDaT5sEVukY0vv8kyZhVYcMZc0GUpa/NsbKxwfz1APIgZss0N2FyazXH1UZnDA4e9832mytsDyQ4SUKDRKSEwVEUmTAemxopMoYTKejGgqyUyQZMCyOSLjOKVvZ7CcBMKrI70aBhBKA50ZBMNAuNWOiwstS89yEMOTyGaVpOpVhtFAXTgEfkTeSXSsuZ5thIR0AawkUq8gtigt0cRKzNq0Mf06QgebY3RCxjMysoOdntdNa0gLpLmJ7NpST0izZ9m2ZdeTVNEQtvSNnxTCSSaTPVp23dafbds9Ssb7xW4G+PQJbw94lnMtIJ5Fe/nyZbLZLO+++y4fffTRgQ3y6Ua7aNn3fU6de42iq/CaYdwTqxSTWZuKH3FvucFUzmFmzaXqhqw1Q04Npbi/2iBtidZ8CYPhZNynu7rWoNHi/alCguFAMb1Up9oisUBp3pnI8fWHZdyulrbXD+W4/7hKpRkghOD4SIrlosvNYmxdTuQc7szXuFOMhQDeOZLn6/dLmBJ+8o+eYiBlQbOCKQU+kqShqUcCS0WEQlKTKUaMkLXIZC5KMe4oBsMIKRQPghRDCUmeEFqZUV+YmOk8IooQQWO9wR/QyTzYCWTgxrVotQ0STKl8XGQcuORUSGglkeVe7bnQTmMkMwjlI7VCic0ZRG06seS8jjDQKMuOj6ULkTBQiSyGjGsLN+YkNBDZaRAGWzUHKGmhWlZdJAyMlsurEbFr27LIlG5Zey2ya19DTyObray/1dVV7ty5g+d5+7b+djPA52Vxrw8Krzzh1Wo1Ll++zIkTJzrKrAc5yKcNrTUffvghAwMDnDt3LtaLI8KLFLVAUqqHVIMI0SLax1WPwaTJlVar2ELVI2vFjfltHBlM89V7xR7V8craGg9LiqAVgJICzo+lufO4RtMLEUKQT1oMJgy+fjcu7TgxnmEgY/HhvVInU/vW4Ty3p9dYdSMSluT4cCYmO6H5x3/2TUZSJlEUYEuNkBKBxDYiakrSDBTjVsRi5ODhMJKMGECx6ErKHhy2EpxwFKBQhkU1hFTCxiGCoE4YxUStnTQkMsjQR0Z+hxjbYw+1k4ZkLm7UD73O+wBGq+5RW0l0MgNRhC0URF21d60SEo2AVB5hWcgoQHQP1emq42tEYGUGMIXG6CLKyHAwW2KdoZUCabS6auJzGcfqghahxSIAnXhdu6hYGLHFuWEqXVzmvf6aUmrXCbRkMsnU1BRTU1NEUUS5XN5k/Q0NDe3Yq/msDvCBF9haBvufTbu0tMTdu3d54403evr9DspybKNer1Ov13uk5YWApCUZSVmsNgLySZOSF5Iy2m6qRcNdN3GytiToug9PD6X42v0S+YTFSt3HlIITg0kuT5c5P5rk1rJLytSkhOYb99cAODORQWnN4mqTW62h2wlLMp61+Le3WyMGpeDCoSwf3I7bsC6dGWKl6nN1pkwmKfgHP/gmQ2kTJQSWahIhsE2oehrfkOQtzVID6naKCUfRCBWPGiYZK67R84MQlKZiZEg7BkboUjAFbbdemw5kh0CFGIHbQ2LQ6oJw0vHYy6AJ/hYuq2lDMgNCIEMPoYItmxI8P8QlQS6ViAVIVbjJbRWhh2ckCJQgk3Va62xYzDCJRBJlmJ25E72LCCLDAcPaRFZSKyJpI0wb0UV2WsdnZKOi8X6rEgzD2NL6u337Np7ndYQ8t7P+DqqP9lXECyU8wzD2FMPTWnP//n1KpRLvv//+pi6N3UpEbYeVlRVu3bpFMpnclPE1pCTnCE4NONwregwlTe4Wm0xkY9HO2YrPZM7BkvCN2QqOKcnYkkO5BB8/WkMD43kHN4woWAaXp+NY12KlyWTOolQLma2tnx/p17m5EHYyuON5Bx0pvnprlYRtkLBijb2v34mb4k+MJok8j+mVJocHE/zk958lnzIwpCSKAlKGpikMFIqEKaiEBgVbM+aErEUWVc9gPKE5akOg4EFNsOYJRjMJ8jKEKCYYZSfBTCB0hIx8dOgioq6BRolMPKtCtd9vojeqlVjJuPcWHdfEKZ9IiM0uq5VAOykkgjQRDsYm2XUAZdhoO4mOIsIgIpPYTDIaEW9jmMgoYCsKiISJkhZSyk2cq4SMCX5DHE/pXje25zMHXIa10fprx/7u3buH4zhbWn+7mVjWz9LuAQc5yCcMQ65cuUIymeTdd9/d8uI5CJdWa82jR49YWlri/fff78QFN14oQggKSYvJnMILY8UUxzK43nJlM7bBVx7FSQYvVLwxlua3bq92Pl+sB+RM0SkvARjJJgkDRbEl526bkpNDST5+UGIkZ1NsRBzOCWaLddwgZr9/78QAN2bK3F6Nb/43D2e4fH8FKeE73zvK//PbD8fSUlJimRA2XQxb4EVgSJNCAlZdzVw1ZNI2GbM0fgQzTYlGMJ6EYznBbR8sCbVIkMjkMHWsl9ctfY4wUKk0wrQQURC/30VwQgi0YaEsB+EkEVphRH5nBkQHlgOtTgjsVDxmUUd0p2KFaXVihVoYBIaD0nH3C1qBFOhksscV1tKMuyhiysNQUSxE0OX+trOvgrhvOkJ0iXlCZNhIw0ZK0SPVvh3ZwcETXjc2TjHbyvobGhra8THUarUDr3R40XjhFt5uiKnRaPDJJ59w9OhRDh06dGDrboRSiuvX43GB77//fqeM4EnBXkPGPayNQPE5K8u9osvp4RS3lhus1D3OjKS5uVTnaCHB1x+VGUiZlBoh41mHctVjqpAAYkJ4bSzFh/fLnBhJx8Oxcw6m1nzUysweHkxzeBi+1kWaJwYEN+8/ZrkZWxpvH8nywe1YmOI//e7z/JH3xhEiFicVAoIwxJQSpCBtCRqRxAg1Q0lJw/VZ8iSGkIwl4UgrhFPxNY9qGtuUKNMha4sOiWhayQLDiV1KLWICC3qtdy3NuEzFMMHxMIMmhBtIrr2elUIbFsK0kCpECPVEPXTlpOPhOCrE3uLh6rkuliVoRALDSWBJgdygVaINCxF6PUTX/aAWMh6mrYSBNp1eoVjRGsuJYF00bGtorZ8Z4W3EVtbfysoKKysrCCFIJpPbxv7q9TpTU1Nbvveq4oXH8HY6uWx5eZnbt2/z+uuvP/Wps5+5Fr7v8/HHHzM6OtpTJP20NU3DYDLrUPcj3h7P8Gv3Vjkx4PD12Qo5x2Qia3NvqY4bKk4MpUlZPssll0ozpNoMGHAEw5kEH96P3dr7y3UunRjgyoMS5UZswiRtA88PuLu4Xmz81pEcX70xD8C5qSSB5/HB7SWStuT/8+e+maMFCynBMQ1CrbGloF73UAo8Jck6gtDTrPmSehTLHx3OSAIFjxsaD0nWloylJNkU1GpxvFFLK5Y+EhIRhQgddfpQddsSEgJtZ8C0QKtWQiGCMEKJ3pteSxNlp0G211MIaSL15pYwLWRr3wYIkIECHW1uHWutm8ykibQiuU2uQAuDyM5sIro2hFatchNz0/tKa7wwHuaktcYwjDixsQWxKaX23OK4H3Rbf/l8nmo1zuTfvn0b3/d7Mr/t4+4nLfaI/bi0bcn35eVl3n///W21uXaz7laoVqtcvnyZs2fPMjw83PPeTkg0aRuMZU0ergX83uODuGHElYUqNT9kMmtTb5WUrDVDIj+i0hrSbUnB8eEMv9tltb0xlWOl2OiQ3VjegSjiG3dXuXR6mNuLdQ7ljA7ZjeQSFBKSr8x6fPfFKf78F86TECGOEZOdFBCGoKKIQUdTDiSVAIQhGUhKckqz0owIhGTWhbGU4HBh/XdbrEckTUEmW0BFAUKFyGhD2xWiZenZMWFFQWxJhVs81IQRl36YNigdi3eqsFckrstXVEZseUGsVyfRoEPQMbF2D98JFYhEOk56aBUrRLfavboRaU3NizCdBLY04sHiG7szhIgtV8PcTLyARqKlxLRiZZEoijrXSRiGHe+gTSLP0qXdKaIoIpFIbGn93b17l0QiwbVr1yiVSnuO4RWLRb7v+76Phw8fcvfu3S8D36u1Lm21bUsP7zrwz7XW/8nev9nT8VK7tFEUceXKFWzb7riWB7HuVlhcXOTevXu89dZbWz7VnkZ47Yt9OGFRskKWmxGWhP/088eZW2vwd353huGMRRhpylWPwVRcmJqxJSlD8sH9EiM5h+WKx1uTWb56K86ynhrPYBqCh/MVqi3Jp3LdI2tEXH0YW4Mnx7Msr5RYKgb8w7/wrRwbS6N9H8eQWIbENKAZKLQQ+K6PlTHJCE3JU6w2FfUABpMSS8J4TtIMNbM1TSNQ2KZkNCUZy8Zko6MIocL4ZjeszlQuoRWoEEHUseZ6zo8w4rm+htEhMsNvwAYZps72CLQ0iJxsvB4qtg5hM/EYNkQBFS/ESSRxbKNFXHp9W2nGld+0yNOwQQiyXWaf63odlZNQ6Vhl2VzPzHZ71BoRk2jbA+hql4SY2NrkF0VR53o8qEL7/WBj0mJj7K/RaPBrv/Zr/O7v/i6/8zu/w3d/93fzx/7YH+Nzn/vcjvfRHuDzxS9+ESHEr7L1AJ82fhT4rT1/oV3ghbu0TyKmdrzu8OHDu44jGIaB7299I21Ed8b34sWLT2xz247wtNadC1lKydmRNAsPS5RdRT1QJE2Dv/pdp1mpuvyzr89xZTVgtR5wbjTB4lrA9Gocv5ssJBhMGHy1y9Iby9n89rXFTn3d+cks92cWOX90iEfL8NaxAoHX5K/9wDfz2tEhUgkT3/VIWaI1kEPE0kZKoPwwLkHxNfmEQdI2qHoRVV8zU1NUGwFRUjCcMjjWZd0preP4mzBjeXXTQqgAoXVcMrIJItaP65BhFGdoibpIp9dS1yKODSINQMezM4gQOnxiREzJmHAjDEwhydnb1KEJEUuwS6MzTGcj7GQKrSICDMzEulhnz3ESk7dGbOlCt7HRqtNa4/s+a2trDA0NEQRBnPl9guv7LKGU2lZ/MpVK8Rf+wl/g448/5kd+5EdYWlraUsZ9O/zCL/wCv/Ebv9H+508Dv8EWhCeEeI941sWXgPd3tZM94IW7tFs98VZXV7l58yYXLlygUCjsen87zdJGUdQZGvSkjG/3mhsJT2uNUqpTTNr9Pb/1aIF/catIzY8IDMFKw+XYQIIf/vfPslj1uDVf5ZOHa1x+FMdS0o7BWs2j2bLihIC3Dmf59U/mODuZ5/Z8jXeP5/ng2gxKa1bLDf6r73+bk2NZDg1nsE2JaUiaTZekAZGOjyVpSRxDEilNPQwIlcYVAg/IWpJs0iTb4ol7fixnfq+scMN48EzeMTiUi+vXBAoReZ1iW4iJKm6xMug010UBUgVPIMN1Symy4yLftnUoiRv2e7aVZkftOLYS7Th2Fw8+RACNeoN8cvODSguJMmyQEhBP5Cfd2o+WJpgO1hYbRpFifmmJWr3JwOAgAwMDO+5ykDKWhrp+/TpTU1MMDg52rL+2BQjx/fA8yG83ZSkjIyO89957u97HTgb4CCEk8BPAnyQWCH3meKlc2nYpyOLiIu+99x6JRGKbT+983a3gui4ff/xxJ47xNGwkvO3IDmKS/33H8/zL20XKKh60/bDk8qioODaU5PNnhvk9Jwf5M992jLlSk/lik393exU/ZbGw5nJ8OMnXbsWZ1lzS4k/93iOgFX/0W76FY8NZxgfTmIYgVKojI95oelgi7odFg2XIOKkgJabUjCQlfquTouopqr5G6wjLlCQtScaxGMqYbJzf1p4foTUIMxFP7tI6jtNpFRNglwS73vhZw4rJpCV/LnSEFq36uW1+p5iwLDSJ1ud013leP9+pVDKO59EmRSv+3vQ+bDeG6GKyNltW39ZE0/7umCZjE5PYpVIn1uU4DiMjIwwPD297rQZBwMcff8zhw4d7uoGAHtLrdn2fpfW3m8Lj7WJ4BzDA5z8G/rXWenY/8m27wUvj0kZRxLVr15BS7nlEYxtPs/DW1ta4du3aroZ8dxPe08iujYxj8vkjOX7pbpGFaoglJIOZBLPlgCuzVc6MZZjIOZw/ZHN6NMvnz46CjNd3gwitY7EBQwrQce7TMuOpWFprIt0id6UJggBbghCxNWeZcXypGUQESrNSrnNqyCEJdOe43UDRDCNqbojvR1xdDIg0mEKQTxiMZMxY7VkIlNZxHdwW51YDSBMtDbTSYDpxIkHHtYnx37us+Q3Xf5zwsFvEGCchhFYgrU72d8ssLCANkwgzzvDyZI+io1PXafrv/e3W1e26SB7ZE6fbGOtaWVnpDKseHBxkZGSEfD7fWbed9T927NiWUmVt19c0zY7r2231RVHUIb6DIr+Dai07gAE+3wx8XgjxHwMZwBZC1LTWX3zqwe0RL4WF17a2Dh06xJEjRw5k3SfF2x4/fsyjR4949913d6Wo0j7WdnKiLdb5tCfTeMZiOCjyIMxSSDus1FyGcwnSGZtbS3UWaz6jaZuhtEXGiclCa0EmEd+8SsejnKWM96XRcT2tITBlTG5eGJA0W4F1FQfqgzCiHqm4q8KLcCzJQi1EKUXCMnBMSdKUJGyThG2yXPU5O7x1o7huHUfT8yEKENLAtB0sy2zlBnTHzW13bgm1dXBeA7REMiMr1bLENGjVIsaN9XabiTHUUG96ZLKZjiWk25/dcp9xAkTLVkxxm99MI1pEt32MDuJY15EjRzhy5AhhGFIsFnn8+DE3btwgk8lQKBSYm5vj1KlTm7L+W6FNaIZhYFlWh/ja11z779uVvewEO80UR1G051kz7QE+X/ziF+EJA3y01p1ht0KIHwTef5ZkBy84hieEIAgCvvGNb/Daa68xMDBwIPvbyqXtlnt/2njGrdC28LqTE0+D53lcvnyZi4cP4dQcPpyvMV5IcONxmclCCidpUYs0MzNrjOfi8YuvHcowkIylw9uWnWpZeqYUBJFGozENiW1KKg0v7tONFH7YUtklHjRkGiaRUriBT0IKGl6EYUgqboAwJFppQq1RSlOt+TT9CEPERHp0OEWiPXGo1RGaNJ24oLhzUrfJWiNAGjHRtNxF0fpMxynVej3zuuUaMZTRjhECWmNKSd7pdSE3dkp0u6ux6kmr2X+LY+6Ul3QEAXfvXpmmyejoKKOjo2itWV1d5dq1a9i2zYMHD6hWqwwPD5PJZHYsHrBV4qO77GWv1t9OLbz9DvD53u/9Xv7+3//7EMfnvhdACPE+8Oe01n9mz4vvAy/UwpuensbzPD7/+c/vOV63FTZaeGEYduSj9iP3vrKyQjab3dGxVqtVrl27xpkzZxgcHGQSWKx63HhcYShtM1uqk3VMMimHQj7J43ITrxkQaMimLJTn89pUnsFk7Ko1vYBsMnavfKUIvIhmEJKxDYJIE2iNEgaRUp32NlsKEpZJ04WRzPZP6rvzmtOjGy28jYKXtJROWuQgBJFSBGFIGMZKLpZlIQWxS936zFYkI9ZXbO1JrBNUK8nQjrr1kOI2v52SZlx+0orJbdxyfUU6D4aY5J5uze0Gruty9+5d3nrrLQqFAr7vs7q6yoMHD6jX6+TzeYaHhxkaGtpV4gM2l73sxfrb6cQyeLKx8jR0DfCBroRE9wCfDfv7h8A/3NPOdoEXQnjt1i2lFKlU6kDJDnpjeO3ylmPHjnVnjXaMdkxlYmKC+fl5rl69Gs+YGP7/t3fm4VHV9/5/zZLJvm8sCRAISyBkUbBqVdRqaTEkaQWUeqvWBerFVr29Vv1pLV1ca2+tWpdWK61eCyaAIASsoq1XARUkG1kIkH2byWTPTGY75/dHPMckJGQymZlM4Lyeh0eTOTPznck57/P97DHExsYSHBx8xklhMBg4deoUS5cuHeL0vXXZTJ77vxpaem2oRDB0W4gJthERFkhgsD8OEfQ9/TS0deGvVWNWqTCbrUSH+hMfpGVxUAR+GhW9/Ta0GhUhflq0auiz2BFEEbtDQKtRERrgR2iAFq1azanWbpJjAhEEEatDwOYYOM7iEHA4RGwOAbtDRFSpOW20EKTTEB8umfrDzMlBaW3SJ9Zq1Gg1OvD/WlBNJhN+/mcK7ICZKImlGjRfByGGNlH6+r3Fr+o2VMMfkV9HM/BaTrgXvvbLqZwyWV3BZDJRXFxMSkqKXBGk0+mYPn0606dPRxAEurq6aGtro7q6Gj8/P2JiYoiJiRlX77mxdn8jJT1LnIvDeZzF6ybt8PkWhw4dcvv7SSZte3s75eXlTpWjjcTg4IROp2POnDnMmTMHq9VKW1sbp06dwmw2D3FW19fX09bWxoUXXjhiTt9PLpvNr/95ks5+B2FBfnSYbTR39hMdHkBMRAA9pn4cX5mCp5s6CAnwo9nYQ1tsGMeaetDYrVyYPI24YD+C/LX09tsHopkqgaAAP8L8NQTptAiCiLHPgtUhcMpoHmgRrwK+CrT4a1X4a9SEBfjRZbIwN36wc3pkU0Y841FJ/QZ2SQNCpMI/OAwBsDvsOL4aaqTz80OjUQ9+1kBiyVnMYol+iw1BFAkIChnIjVOpz8ilExkhCjvofWQ59WA0sLe3l5KSElJTU0dtq6RWq4mMjJTdN2azWe7IY7FYiIqKIiYmZkiJ11iMtvsbnvQs+Tud8eHZbDa3jl7wFby6w5Oio4sWLZIjXZ6YXKbRaDCbzVRVVbmc3nK24IROp2PGjBnMmDFDFtampiaKiorQarUkJyeP+nlUKhWPXDuPn2wvo9FiIzRAR2CQH21dZpoM3cxNiMLY3ktIiIY+B5hMdlSimsrGDvq6ewkM8Kexy0yfyYJarSZlZgQxQTqS48OYFh+Gv1ZDe5+Fjj4rXf0WQnVaVKiYERlERKB2yOwFiVqjeSB1ZPAWbuQvBeGrU2ako+TfiQJqUUCn1cJZfKV2x0ArfDncoRoo05IEVPo7CFoIGiPA9LWZ6j2BG0xPTw+lpaUsXbp0XPWngYGBJCYmkpiYKJ9Lra2tVFZWEhQURGxsLNHR0U6VVEoM3/0N/yeJ4dlM376+vnOu2zF4UfCkebSZmZlDvkgpNcVdgicIAhUVFdjtdpYtW+bS1n2w2I21Lo1GQ0REBPX19cyePZvIyEgMBgM1NTVynlZsbOyQE1ajVvO77EXc/XYJLWYbEYE6/IN0mM12jle3MS08gBZ9J9Ex4Zh7zWgD/ekwdhMRE4Optw99m4nwyHDsViv7ihpRadSEhQaj0miYFx3E3Kgg5kUFoNOomT8tQn5fu0PA0Guhy2THZLXhsAkDU8osNvosdoL8h54Ow6ViINXXOUf20F2gasD3p/rq/xGx2ez099vQ6XT4+/vL3/MZfjeVioBRxG7wDk7+eZR8Ok/S1dVFeXk5aWlpE+ofp9Fo5PNFFEX6+vpoa2ujpKQEQRCIjo4mJiaGsLAwlwIfDoeDEydOEB0dLVsvku91eNJzb2/vOdc4ALwkeKIoYrFYuOiii84QIHdOLrNarRQVFREdHU1AQIBLYufM3W8wJpOJkpISkpKS5DyriIgI5s+fj8lkQq/XU1JScobfL9hfyx++v4T7dpTR2mPB0dbHjGmhOBxqqlp6CAn1x97STkBQAG3NRqLjo+lo60BUawiNCKO/pweVWkN4TCQIAt0d3fhpVXyhb+MIKnpEP0S1hqgQf6JFC0nTIpkT5U9CZAhz4kKZP+1rk8tssRKoO/NUEIZo28CAhtHlbtAFKApfVWaM+OhAUMVfg5+//xmDrEd7ZWnnxqD/emv3djY6OzupqKggPT3drYOjVCoVISEhhISEMGfOHGw2G0ajkfr6enp6eggLCyMmJoaoqCinrh1pyJVarWbBggWyZTVa0vO5KniqMULPrselh2G1WkcMcxcVFTFv3rwJf7nSbIvk5GTi4uI4ePAgl156qdPPdzaZeDCdnZ2Ul5ezePHiMX2Ekt/PYDAM8fvZNYH851vFmFQqVDY7QToVQWHB2Exm7Dp/OvRGIsKCsZv6iI4Mpd9qQ6sCTWAQlv5+NA47bZ19+Ov86LapcDgEtCoHKgbmLnR09uEfHAYWE/12FaKlH7Vaiz8C8+JDuXPlQi5ZMvPsJqr0HY3wu9G+y+FTvkY8jqHJvgAWq5XGxiZsNht2h4OgoGAio6IIDQ11OWLoKdrb2zlx4gQZGRluD7ydDVEU6e7uxmAw0N7eLhf/x8bGEhQUdGaX6K9SskRRZOHChSN+j8OTnvPy8tiyZYu7fOw+84fzmknryUE+BoOBqqqqM2ZbOIsrYtfc3Ex9fT2ZmZlOnewj+f2am5vp6uriv5YH88fDfRjtAlZRQ7++E1GrQejtJzgqEqPBSFBkJPWGbtT+A9Fcu751oKOwfyAabQDtxi60Oj8CQwbmXth7eum3CoQE6OjrbMeuC0NlNaMSwWEXsApWLlkUxSVLBsrqRpIncYTfjiZjw78xYYRzfHgcVpa8r77vgYDWwECm2NhYeVdTW1srd9+NjY0lKipq0qOMRqORkydPkpmZOS7/mjtQqVSEh4fLN1mLxSKXu5nNZiIiIoiJiSEyMhK1Wk1VVRWCILBo0aJRz+3BgY/y8nKef/55/vKXv3jtM3kLr+3wbDbbiNUP5eXlxMfHO13iNRipV15bWxvp6elDssKd3eGNt3JC6q7S09NDamrqhJs5SnfrxuYWNu9rxmB2oAoMxNZnIjhIh8NuRRcehmg2QWAw/d3dBPgPjAnsbW8n2N8Pu1aHVqPF2tVBT3cf/gGBCNpANCqB/p5eHDYHGq0fWpWKfnM/foKF//5+Jjdcs3RCa3cnfX19lJSUsHDhwhET0KV0DmlX42wdqycwGAxUV1eTkZHhciWCpxAEgY6v6n07Ojqw2+34+/uTmprqlMldVlbGj370I/Ly8li0aJG7luUzO7xJF7yqqirCw8NHrDM8G4IgyLW3KSkpZ/jbDh48yCWXXHL2MqJxip3D4aCsrAydTif7QdyJKIr84u0iPi7Xo/X3wyYM5MfpVHasVhsWSz9R0eH09fQR4O+HWqPFaOzEX63CJmjQBkegFiz09/aiRsSuCkQQ7Kit/QM7bJuFQI3Ak3dewRXpSW5d+0To7u7m+PHjZ03nGI7JZMJgMNDW1obD4ZBNOk+bvq2trdTV1ZGRkeHTaRuiKHLq1ClMJhPh4eEYjcZR630lKisrueWWW3jrrbdITU1153LOP8Gz2+0jmq6nT58mMDBwXEnBFouFwsJCpk+fPmrt7eHDh1m+fPlZR9WNx4SVAiLTp0/3eJ//A8VN/DL/OA67Df9AP/osdrR+fqhVKvq6uggMjwKLaUDUtIHYrf1gNaP1D8Bus4CgAYcVu82BXeVHoGDG5nAQEyDyyv3XkZwQ69H1j4eOjg4qKytJS0tzOQ1CMn0NBoNHTd/m5mYaGxvJyMiYlDbt4+HUqVP09/ezePFi+fyW6n3b2tro6uoiJCSEsLAwAgMD6enp4aabbuLvf/87GRkZ7l6OzwjepP/VxuvD6+7upqSkZEgu39led/gJ74q/rre3l9LSUubPn3/W93QX30qbwTcXxvLQG0c4UNKKn84Pm7kbQaMhKFCHra8LdWAYWHuhvwc/VJgsVqwmCw5tMH4aEXNXNyqNDj+xD6ujn2syEnj67ut86kLV6/VUV1dP2A/m5+fHtGnTmDZt2hDTVxpV6A7Tt6GhAb1eT2Zm5qT7D8dCSohfsmTJkPN7eL1vT08PRUVF/OxnP8NgMLBmzRrUarVs8ZyLTPrZr9VqsdlGbhQ5nJaWFk6fPk1GRsaY+U7jbdg5GpIzODU11ath+gB/P/5wxyUcKmvkF298gbFfjcoBnf0iOsGCvc+IXdSgtfdiwx+Nfwg6oRvMbfTZ1QRiw9ZvIjE2kKfuyiVtwUyvrd0ZmpqaaGpq4oILLnCraTi8kkEyfY8fP+6y6VtXV4fRaCQ9Pd3nxe706dMjit1wVCoVYWFhJCUlodVqef311zEajTz++OPce++942rnPpXwmknrcDhG7G7c0tJCX18f8+bNG30RX/kjurq6SEtLc+oCKSwsZP78+bIwjtdfB1BfX09raytpaWmT7pze/UkFfy4opqXLhsncj90hEBwcNGDOigI9Zjt+WEEQ0OJg/swINq25lJWXLJ7UdY9EbW0t7e3tpKWleVVAXDF9a2pq6OrqYunSpZM+fGcsqqur6e3tJTU11elMgzVr1vDHP/6RK664wpNL85nt4qQLnsFgoKOjgwULFoz4PLvdTmlpKQEBAaPmEI1ESUkJc+bMITQ01KVIbGVlJXa7ncWLF/vUiW622Nj9SQXHqpqobmqnz9RPZ08ffogsSopnReZcLlsyDaPRiNlsJjIykri4OMLDwyf9c0g3LmkHMpnrGSvqK4qi3N1kstfqDDU1NXR3d5OamurUWltaWlizZg3PPPMMV199taeXpwieREdHBy0tLaSkpJzxmNlsprCwkFmzZjFz5vhMsuPHjzNz5kzCwsKGZJCPhd1up6SkhPDwcJKSknzalyEIAqWlpQQGBpKcnHzGWqV8P4PBQFdXF6GhoXJtprd9eaIoUl5ePjDkaBw3Lm8xPOoLA77B9PR0nxe72tpaurq6nBY7g8HA9ddfz2OPPcbKlSu9sMLzUPAEQRjRV9fd3U1tbS1Llw7NCevo6KCsrMzlxqAVFRXyYGFnTViz2UxxcTGzZ8+WZw/4KjabjeLiYuLi4khMTBzz+MHZ+W1tbaPW+XoCSZiDg4OZO3euz4ndYERRpKKiApPJhE6n87mE5+HU1tbS2dnptMltNBq5/vrr+eUvf8l1113nhRUCPiR4kx60GClKKzUaGG8bdglRFNFqtTQ0NAA4JZhdXV2UlZWRkpLi0qQ0byIJ8+D63bEYnJ2fnJws72hGqvN1pyBJzVdjYmLc0r7fk0hip1arueCCC+R6U09Efd1BXV0dHR0dpKWlOSV2nZ2drF27lv/3//6fN8XOp5j0HV5/fz/Hjx/nwgsvlH1nZrOZpUuXumR2De4C0dHRIZtz0l06Ojr6jJOjtbWVmpoa0tLS3FoA7gmkNkTuFObhdb6RkZHExsaOqyfbaK9bVFREQkKCS81XvYkoipSVleHv78+8efNGFf3JSngeTn19PUaj0Wmx6+7uZs2aNdxzzz2sXbvWCyscgs/s8LwmeNIg4uHY7XaOHj3KBRdcQHFxMeHh4Wc94cZ6j5GCE6Io0tnZiV6vp729neDgYOLi4oiOjqahoUE2CXwpR20kjEajXDM8kTZEZ8Ndfj+p0atUF+vLSCZ3SEgIc+fOdfp53kp4Hk5DQwMGg8Fp/2Jvby9r165lw4YN3HTTTWMe7wEUwRv8+08//RS1Ws3cuXNd9p2JoujUgB1RFOnt7aW1tZWGhgY0Go08Qs/bReDjoampicbGxjNqhj2Jq34/qS520aJFPu8eEARBDlLNmTNnQq/jjVpfSeycTekxmUysW7eOm2++mVtvvdVt6xgniuBJGI1Gjh49ysUXX0xYWJhLrzveZGKr1So7/GNiYjAYDBgMBkRRJDY2lri4OJ/p9io1SOjs7PR63tpwJHPOYDAgCIIsfoP9fq7UxU4WDodD9i86E/gZD54wfRsbG2ltbXU6AdpsNrN+/XrWrFnDnXfeOZnBIkXwYMDp2tzcjM1m47LLLnPpNccrdtLuY6RZoVarFb1ej8FgwGq1EhMTQ1xc3LhG67kTQRCorKxEFEUWLVrkU+kRg/1+JpOJqKgoAgICaGpqIj093WduGKPhcDgoLCxk2rRp4055Gi+S6dvW1kZPT49Lpm9TUxMtLS1Oi53FYuGmm25i1apVbNq0abIj4+ef4MHAHwEGLuTy8nIEQWDJkiUcPnx4XM06wTWxkxo2OlMmZrfbaWtrQ6/X09fXR1RUFHFxcXKai6dxOByUlJTI5T++nMrhcDiorq6moaEBnU5HWFjYpOX7OYPdbqewsJCZM2d6PZjiiunb1NREc3MzGRkZTomd1WrllltuYcWKFdx3332+cO5M+gIkvCp4VqsVi8VCUVERMTExzJkzB5VK5VJ34vGWiTU0NNDc3ExaWtq4fXWCINDe3o5er6erq4uwsDDi4uI85pyWopszZ85kxowZbn99dyPVxaanp6PVamW/n9FoRKfTeS3fzxlsNpuczB4fHz/ZyxnT9G1ubqapqclpsbPZbNx2220sX76cBx54wBfEDs5XwTMajRQXFzN//vwhkbtDhw7xjW98wymTzdngxODjq6qq6O/vZ8mSJRMWKFEU6erqQq/XYzQaCQ4Olu/Q7iiCl+aaeqszy0QZqy7WGb+ft7BarRQWFpKUlOSTkePhpq9Op8NisbBs2TKnAlV2u52NGzeyaNEiHn30Ubd8v7fddht79uwhLi6O0tLSMx4XRZF77rmHgoICgoKC2LJlCxdccMHww84/wRNFkc8//5zk5OQzzMkvvvjCqYaK4+1hJ5mFISEhLqe6nA0p4iuJn9R+x9XdjJT8PBUc/tJQGOlG4szNZyS/nzvy/ZxB6qGYnJw8JW4kLS0tVFdXExkZSWdn55imr8PhYNOmTSQmJvLb3/7Wbef6xx9/TEhICDfffPOIgldQUMDzzz9PQUEBn332Gffccw+fffbZ8MPOP8GD0bsef/nll6SkpIya9OuKv66/v5/i4mISEhK8ZhaazWY56DHeiK/BYOD06dNTIvnZHXWxDoeDjo4O2U3gyTrf/v5+CgsLWbBggUujBLyN1FU5MzNT/i5GM32lzcO9995LREQETz/9tNtvHjU1NWRlZY0oeBs3buTKK69k/fr1ACxcuJB//etfw32jPiN4PuFRPlsTUFfETkqNWLRokUt1uK4SGBjI7NmzmT17NlarFYPBQGVlJVarlejoaOLi4kZMS2hoaKClpcXtveE8gZS3FhoaOqFgikajISYmhpiYmCH5fjU1Nfj5+ckpQxPNYTObzRQVFU2JnEAYaIo6XOwAgoKC5HNLMn1Pnz7Nj370IyIiIoiLi+PZZ5/1eiS/sbFxSEpPQkICjY2NPltZ49OC50pwQuqiO9mpETqdjpkzZzJz5kzsdvuQ6VuDTbnTp09jMpmmRCddqS42NjbWrXlro9X5Sk07pfSg8fr9pBSkxYsXu5Tj6W30ej21tbVjtpCXOjzHxcXxne98h8bGRpKSklixYgUbNmxg48aNXlz11MInBE+r1Z7ROsqV4ITkQPe1nZJWqyU+Pp74+Hg54tvS0kJRURH+/v4kJydP9hLHxJt1scN3M21tbfJAGmf9fr29vZSUlEwJfygg724zMzOdHqz961//mu7ubrZt2ybfLEdK7vckM2fOpL6+Xv65oaHB43mNE8Grgjfa3Xn4Dm+8wQlBEKioqAAgIyPDpxJ0h6NWq4mIiKCuro6kpCQiIiLQ6/XyMCPJlPMlwZbqYufNm3dGsran8fPzY/r06UyfPl2+WbS2tlJZWTmq309qsJCWluaxmmN3Io19HI/YPfHEEzQ3N7Nly5YhloG3O3NnZ2fzwgsvcOONN/LZZ58RHh7us+Ys+MgOTxI8V/x1Ul84qf2Qj+QdjYoUTJk9e7acBxYREYEoivT19aHX6zl27NiEI77uoq+vj+LiYp9om6VWq8f0+/n7+3Pq1KlJd2k4S1tbmzzj1lmx+/3vf09VVRX/+7//63E3yPr16/nXv/5FW1sbCQkJ/OpXv5K7Hv34xz9m1apVFBQUkJycTFBQEK+//rpH1zNRvBqlHW1UY11dHSqVihkzZuBwOFCr1U4Jl5SzNnfu3HHPtZ0MJDPLmWDKRCK+7mIq1cWaTCbq6upoamoiKChIns41Gfl+zmI0Gjl58iSZmZlO7cxEUeT555/n888/Z9u2bT5lBYyBz/wBvCp4o7V5b2hooL+/n9mzZzu9s+vo6KCiomJKXIwwUNZWWVnJ0qVLxz39TIr46vX6MSO+7lzviRMnJjQv1pu0t7dTVVUl15pKZYHezvdzFlfE7pVXXuGjjz5i+/btkz5Uapwogie/wVfJuxUVFdhsNmJiYoiPjz/rnbmpqYmGhgbS0tImveusM7S0tFBXV0d6evqEzVMp4qvX68+I+LrrYtbr9dTU1Lhlvd5ACmqMJB6S389gMNDZ2Tmpcz0kJHEej9j99a9/Ze/evbzzzjtT4pwfhiJ40s+D/XVSRE6v12M2m+V0BGknI0296uvrIzU11efTOAZHjtPS0tx+gXmixrexsZHm5mbS09OnhMkkiXNGRsaY4jHY72c0GvHz85NL3bwlItLOeTzDx//+97+Tn5/Pu+++6/NJ6aNwfgqe1ObdmeCEw+GQxa+3t5fIyEh6e3sJCwtj/vz5PuuXkZDa1TscDlJSUjxuSg2u8W1vb3cp4ltTUyPPSPD1mwkM7Jzr6+uddvgPx2w2y3W+Ur6fVL3gifOro6ODysrKcYnd1q1beeONN9izZ8+UiDiPgs9crF4XPKvVOu5IrNlsliOXDodDnrUaGRnpk8LncDiGtAz39hoHR3zb2trQarVy0GOkC02qi7VYLD43h3c0pJZJUoeWiSJZFwaDQW4H5k5XQWdnJxUVFWRkZDi9m9yxYwd//vOf2bt375TwU58Fn7lIvSp4XV1dcgTWWbHr6enh+PHjch2kIAh0dHTQ2toqD+eRzDhfuFClbsrTpk0jISFhspcDfL2T0ev1csRX6lgi1cVqNBoWLFjgkzeQ4TQ0NKDX651uhjle3O33c0Xs3n33XZ577jn27t076elAbsBnTiqvCt5LL73ESy+9xFVXXUVubi7Lly8/q0hJo/FGG1ojDedpbW2lo6OD0NBQeTjPZJhkUt3mvHnzfLL9EAyN+FosFhwOB1FRUSxcuNAnbhhjUVdXR3t7O0uXLvXK33iifr+uri7Ky8vHJXb79+/n6aefZu/evVOis4sTnJ+CBwOi8M9//pP8/HwKCwu5/PLLyc3N5ZJLLpFPYFEUqa+vx2AwsHTpUqcjWd3d3bS2tsp96iQfljeicVLO2uLFiwkPD/f4+00UqetvYGAggiDIflKpq7Mvil91dTXd3d1OD532BOPx+0lil56e7nSw4cCBA/z617+moKDAZ2+aLnD+Ct5gLBYLBw4cIC8vjy+++IJLL72U6667jq1bt3L99dfzne98x6UTe/Bksra2NgICAuSqBU9EHtva2jh58uSUyVmT6mITExPlKXGSq0Cv19PZ2TmkTftkBzBEUZSbLDjbe88bnM3v19vbS1lZ2bjE7uOPP+bhhx9m7969Lk/vG87+/fu55557cDgc3HHHHTz44INDHt+yZQv333+/XP969913c8cdd7jlvQehCN5wbDYbe/fu5ac//SkxMTGkpaWRm5vLlVdeOeEkS6lJp+TAl7Lw3ZG82djYKLc3nwrJoFJvuJGGGElIEV/JjJvMGl8poGK1Wlm8eLHP+hgH+/2MRiM2m43k5GSmT5/ulIXx6aef8vOf/5y9e/e6rX+jw+FgwYIFvP/++yQkJLB8+XL+8Y9/sHjxYvmYLVu2cOTIEV544QW3vOco+MwfzSdqaWGgTvL3v/89f/jDH8jJyeGTTz4hPz+fX/ziF6Snp5OTk8O3vvUtl/KlQkJC5IipyWRCr9dTVFSEWq2Wd37jfV1p19HT08MFF1ww6bsgZ3C2LlalUhEREUFERATJycln1Ph6K3dNSu0BfFrs4Os6X39/fzo7O5k/fz49PT0cPXp0TL/fZ599xv3338+7777r1ma1Uodxabj4jTfeyK5du4YI3vmGz+zwADnqOhiHw8GhQ4fYvn07Bw4cYNGiReTm5vLtb397wuZjf38/er1ejl5KO7+xTBBp6ppGo3G546+3kdrHu1LaNpjBPixBEIb0qnMnUvRYq9VOibxL+LpLy/DGBSP5/cLCwoiKiqKwsJBNmzaxa9euCQ0CH4n8/Hz279/Pq6++CsAbb7zBZ599NmQ3t2XLFh566CFiY2NZsGABf/jDH9w+oxdlhzcyIzn7NRoNl112GZdddhmCIHD06FHy8vJ4+umnmTt3Ljk5OXznO99xKU8pICCAWbNmMWvWLHkmbXl5OXa7Xc5bG34h2+12SkpKiIyMlGt/fR0puz8jI2PCmfqBgYFDvrO2tjZ5SJLkwA8LC5vQ9yIIAmVlZQQEBHhkFokn6O3tlVtSDb8RD/7OJL/f3/72N15//XVsNhuPP/74pKUwrV69mvXr1+Pv788rr7zCLbfcwocffjgpa/EGPrXDGw+CIFBcXExeXh779+9nxowZ5OTksGrVqgnnLdlsNgwGA62trUMGcvv5+VFcXExiYqJP9/wajLfqYqXKGIPBQE9Pj8sRX0EQKC0tlVvITwWkLjjj6b9XVlbG7bffzt13382xY8f45JNPOHDggFsjs4cOHWLz5s289957ADzxxBMAPPTQQyMeL6UodXV1uW0NX+Ezd6wpK3iDEUWR48ePk5+fT0FBAVFRUeTk5JCVlTXhPCZpIHdTUxMdHR3ExcUxa9asCe9ivMFk1cUOj/g6mx8p3cSk3fNUQPKLjsdVUFFRwa233spbb71FamoqgDzGwJ3Y7XYWLFjAgQMHmDlzJsuXL+ett95iyZIl8jHNzc3yzXvnzp089dRTHD582K3rQBE8zyGKIidOnCA/P589e/YQFBRETk4Oq1evJi4uzqWTqrOzk/LyclJSUrBarbS2tsqdSqRdjK+Jn6/UxUr5kdIoy9Eivg6Hg6KiIrfPy/AkrojdyZMn+Y//+A/eeOMN0tPTPbzCgTGK9957Lw6Hg9tuu42HH36YRx99lGXLlpGdnc1DDz3E7t270Wq1REVF8dJLL7Fo0SJ3L8NnLo5zTvAGI4oi1dXVbN++nXfeeQc/Pz9Wr15NTk4O06dPd0qkWltbZZNwcIRNEAS5TVN3d7c8OSoyMnJS88R8uS5WqvGVHPgajUYuC6yoqGDatGk+PQ9hMK6IXU1NDevXr+e1115j2bJlHl6hT6EInrcRRZGGhga2b9/Ozp07cTgcZGVlkZubS2Ji4ojiV1dXh8FgIC0t7awmoSAIcomblLQrmXDeFBwpeqzVaqdEXWx/fz/Nzc3U1NSg0+mYMWOGRyK+7sZkMlFUVDSu5rP19fWsW7eOV155hYsvvtjDK/Q5fOZEPG8EbzCiKNLS0sKOHTvYsWMHJpOJ6667jpycHObOnYsgCBw+fJiwsLBxZ/ZLSbutra20t7cTEhJCfHy8xysWpA4tE50X601sNhuFhYXMnj2biIgIuR2YOyO+7sYVsWtqamLt2rU899xzXH755R5eoU/iM3/A81LwhmMwGNi5cyfbt2+nra0NtVrNhRdeyDPPPDOhHdpI/qv4+Hi31/fa7XaKioqIi4ubMv4vq9VKYWEhSUlJZ0Qm3RXxdTdSc4jxzLltaWlhzZo1PPPMM1x99dUeXqHPogieL9LZ2Ul2djazZ8+ms7OTxsZGVq5cyfe+970J+8Ok+l6pxE2n0xEfHz/h+l5JOGbNmuW2+ktPY7FY5PK2saLorkZ83Y0rYmcwGPj+97/PE088wbe//W0Pr9CnUQTPF3nzzTcJDQ0lJycHGKhOePfdd9mxYwenT5/m2muvJTc3l/T09AnvNKRyLYPB4HJ9r3QRnq0u1teQ1rxw4cIxJ7cNZ6Qds1Sy5cm0G2nNKSkpTnfCMRqNXH/99fzyl7/kuuuu89japgiK4E01ent7KSgoID8/n4qKCq6++mpycnLG7OnnDNJIRr1ej0qlksXvbLWqvjQv1lmksZrjEY6z0dvbe0bE1901vlKzhfGsubOzk+9///s8+OCD5Obmum0tUxhF8KYyZrOZ9957j/z8fIqKirjiiivIyckZ0tPPVfr7++UGnYIgjDiP1l11sd5EEuglS5Y4bRKOB6kuWqpXHa00cLyvWVhYyKJFi5y+qXR3d7NmzRruuece1q5d6/J7n2MogneuYLFY+OCDD8jLy+PIkSNceuml5Obm8s1vfnPCZpbUnbi1tRWbzUZsbCw6nY6GhoZx9VmbbKTSK28JtFQaKEV8pTm+44n4WiwWjh07Ni7Tu7e3l7Vr17Jx40Z+8IMfTOQjnGsogncuYrPZ+Oijj8jPz+fgwYMsX76c3NxcVqxYMeFeeTabjVOnTtHc3ExAQACxsbHEx8d7bMKWu5A6QY+nztSdOBwOOUFcivjGxsaeNUHcFbEzmUysW7eOm2++mVtvvdWNn+CcwGdOUK8IXl5eHps3b6a8vJzPP/981CzzsbqzTiXsdjv/93//R35+Ph9//DHp6enk5uZy9dVXu+RjamhooLW1lbS0NFQqFUajkdbWVkwmE1FRUcTHx/tczprU4txXOkFLEV+DwTDqDBQpgjx//nyioqKcel2z2cz69etZs2YNGzZs8ORHmKr4zEnpFcErLy9HrVazceNGnnnmmREFz5nurFMVh8PBwYMH5Z5+ixcvJjc3l2uvvdYpIaipqaGzs3PEwTUj7WDi4+Mnvb5XmsE6nuE13mR4xDcgIICoqCiamprkCXnOYLFYuOmmm1i1ahWbNm1y23c+1s3fYrFw8803c/ToUaKjo9m2bZvb++m5EZ8RPK/0w0tJSRnzmHO5O6tGo+Hyyy/n8ssvRxAEjhw5Ql5eHk8++STJyclkZ2eP2NNPFEWqqqqwWq2kpaWNaIJJ0cm4uDi5zXhzczMVFRWEh4cTHx/v9fpeo9HIyZMnxzVw2tuoVCrCw8MJDw9n/vz5dHZ2UlxcjFarpbq6mt7e3jEj5VarlVtvvZVrrrnGrWLncDjYtGnTkJt/dnb2kGvhtddeIzIykpMnT7J161YeeOABtm3b5pb3P5fxmQagjY2NQ6oEEhIS+OyzzyZxRZ5BrVZz0UUXcdFFF/HUU09RVFREXl4ef/zjH0lISCA7O5tVq1YRFBTEtm3buOiii1iyZIlTF5PUZjwmJgZRFOWE3RMnTsj1vVFRUR5N2DUYDFRXV5OZmTklZnzAgHCdOHGCJUuWEB0dLUfKjx8/Lkd8pclkEjabjdtvv51LLrmE++67z627aWdu/rt27WLz5s0ArFmzhrvvvtsjLabONdwmeNdccw0tLS1n/P6xxx6TE3kVhqJWq8nMzCQzM5PHHnuM0tJS8vPzWb16Nd3d3WRmZvLd737XpZNYpVIRFRVFVFSUXN+r1+s5efIkwcHBcombO8VPr9dTW1tLZmam14f9uIpUzzt37ly56iMgIIDExEQSExPliO+pU6cwm82cOnWKadOm8eabb5Kens4DDzzgdpFx5uY/+BitVkt4eDhGo3HKJKBPFm4TvA8++GBCz585cyb19fXyzw0NDVOmVZA7UKlULF26lNmzZ3Pw4EFWr16NVqtl3bp1hIaGkp2dzerVq4mNjR33BTZ4KI8oivT09KDX66murnbbRLLm5mYaGxvJyMiYUmJ37NgxkpKSRhUKPz8/ZsyYwYwZM3A4HLS0tPCb3/yG2tpaoqOjOXDgAFddddWUGOKkAD7TLG358uVUVVVRXV2N1Wpl69atZGdnT/ayvE59fT0bNmxg8+bNPPLIIxw6dIiXX34Zs9nMD37wA7Kysnj55Zdpbm5mjIDTiKhUKsLCwkhOTubiiy9m3rx5mM1mvvzyS44dO0ZjYyNWq3Vcr9nU1ERTU9OUFTtn26qrVCo+/PBDVqxYQV1dHddffz379+93+9qcufkPPsZut9PV1TXh7t7nA16J0u7cuZOf/OQnGAwGIiIiyMjI4L333qOpqYk77riDgoICYOTurApfI4oidXV17Nixg507dyIIAqtXryY3N5eEhIQJm1bSCEu9Xj8kGHK2wEN9fT0Gg4H09PQps8sZ3JYqLi7OqecIgsD999+Pn58fzz77rEeDQM60Zv/Tn/5ESUkJL7/8Mlu3bmXHjh28/fbbHlvTBPEZx6KSeDxFEUWR5uZmuaef2Wwe0tNvouI3eIQlIJdqDa7uqK2tldvI+1Jn5bNht9s5duwYs2bNIj4+3qnnCILAww8/jMVi4cUXX/TKZx2rNXt/fz8//OEPOXbsGFFRUWzdulUOcvggiuApuBe9Xs/OnTvZsWMHHR0dfPe73yU3N9ctnY8tFossflLU0mq1YrFYSE1NnVJiV1hYSGJiotNiJ4oimzdvxmg08pe//GXK7GJ9DEXwFDyH0Whk165d7Nixg+bmZrmnX0pKyoTFyWKxUFZWRk9PD/7+/sTExBAfH09wcLBPp0RIYpeQkOB030BRFHniiSeora1ly5Ytiti5js+cGOeN4LW3t3PDDTdQU1PDnDlzePvtt0esk9RoNCxduhSAWbNmsXv3bm8v1a10dnbKPf1qamq45pprXO7pJyVC2+12UlJScDgccpG+2WwmOjqa+Ph4QkNDfUr8HA4HhYWFzJgxw+l5wqIo8vvf/56ysjLefPNNt3ao9iaiKHL55Zfz8MMP893vfhcYKPV87bXXPBJwGQWfORnOG8H7+c9/TlRUFA8++CBPPvkkHR0dPPXUU2ccFxISQm9v7ySs0PP09PTIPf0qKyv51re+RU5ODsuWLRtT/ERRpLKyEoCFCxeeIWhSW3a9Xi+PsIyPjyc8PHxSxc9VsXv++ef5/PPP2bZt25SJPI9GaWkpa9eu5dixY9jtdjIzM9m/fz/z5s3z1hIUwfM2Cxcu5F//+hfTp0+nubmZK6+8Ur6AB3MuC95gzGYz+/btY/v27RQXF7NixQpycnK4+OKLzzDdRFGkrKwMnU5HcnLymAI2fITlZM2kkMRu+vTpzJgxw6nniKLIK6+8wkcffcT27dunTLXIWPz85z8nODiYvr4+QkND+cUvfuHNt1cEz9tERETQ2dkJDJzUkZGR8s+D0Wq1ZGRkoNVqz5uOtRaLhffff5+8vDyOHj3KpZdeyve+9z2++c1vIggCe/bsIT093aXo7/CZFOHh4XKJmyfFTxrsHR8f73QCuyiK/PWvf6WgoICdO3f6ZNMDV+nr6+OCCy5Ap9Nx5MgRb9c4+4zgTU3HxCicrbxtMCqVatQLt7a2lpkzZ3L69Gmuvvpqli5d6s2t/6Tg7+9PVlYWWVlZWK1Wuaffz372MzQaDcuWLSMrK8sl01StVhMdHU10dDSiKNLZ2Yler6eqqspjA3lcETuAN954g927d7N79+5zSuwAgoODueGGGwgJCfHZhg7e4JwSvLOVt8XHx9Pc3CybtKMlnEoXyNy5c7nyyis5duzYOS94g9HpdKxcuZIVK1awbt06EhMTEUWRyy67jMzMTHJzc7nqqqtcEgSVSkVkZCSRkZFD2jOdPn2aoKAgucRtIgECQRAoLi4mLi5uXGK3detWtm3bxp49e6ZMJ+nxolarp0wKkac4bz59dnY2f/vb3wD429/+NmJDg46ODiwWCwBtbW18+umn50R7Klc4duwYWVlZ/OlPf+LFF1+kqKiIO++8k48//pgVK1Zw2223sWvXLkwmk0uvL7Vnmj9/Pt/4xjdISkqit7eXI0eOUFhYSFNTEzabbVyvKQgCRUVFxMTEkJCQ4PTztm/fzpYtW9i9e/ekdGVW8B7njQ/PaDSybt066urqmD17Nm+//TZRUVEcOXKEl19+mVdffZWDBw+yceNG1Go1giBw7733cvvtt0/20n0OQRD4/PPPyc/P5/333yc5OZnc3FxWrlzplpkVfX19tLa20tbW5vQIS2lnFx0dPa5h5Lt37+b5559n7969U2b6m6ts3ryZkJAQ/vu//9vbb+0zPrzzRvAUPIMgCBQWFpKXl8f+/ftJTEwkJyeHVatWuWUUo1TfazAYUKvVconbYJNaEARKSkqIjIxk1qxZTr/2vn37+N3vfkdBQYHTHY4VXEIRPIVzD1EUKS0tJS8vj4KCAmJjY8nJySErK8stgjK4vlcURbkx58mTJ4mIiGD27NlOv9YHH3zAb37zG3mdnuB8TXYfAUXwFM5tRFGkoqKC/Px89uzZQ3h4ONnZ2WRlZbnU0284VquV1tZWTp8+jVqtJiEhwek5tP/+97955JFHKCgocLqm1hWUZHcZRfDOJ86xgSzjRhRFTp06xfbt29m1axf+/v6sXr2anJwcpk2b5pL4CYJAaWkpoaGhJCQkyPN7LRaLbPaONMLy008/5YEHHmDPnj1OJyO7ipLsLqMI3vmCM9PYXnzxRYqLi+XeZjt37jxnB7KIokhtba3c0w8gKytrXD39JNM5JCSEpKSkIY/Z7Xa5xM1kMhEdHU1ISAjx8fF88cUX3Hfffbz77rvjCmy4ipLsLqMI3vnCoUOH2Lx5M++99x4ATzzxBAAPPfSQfMzKlSvZvHkzl1xyCXa7nWnTpmEwGHyqAN8TSD39tm/fzo4dO+jv7ycrK4ucnBySkpJG/PyiKHL8+HGCgoLG7P8mjbB87bXXePPNN7Hb7fzP//wP119/vdvy0c6W7H7LLbcMEbjIyEg6OjrOOLaxsXFIsvuBAwfOtdxPnzmRz5s8vMlipIEsjY2Nox4zeCDLuY5KpWLGjBn85Cc/4cMPP+Sdd94hOjqa//qv/+Kqq67i6aefprKyUm5l73A4KC0tJTAw0Klml1LX5lWrVhEWFsajjz7Ke++9R0ZGBnV1dW75DB988AGlpaVn/MvJyZGT3YFxJ7sreAZF8BR8ApVKRXx8PD/+8Y/55z//yb59+0hMTOSRRx7hiiuu4Le//S1r167l8OHD4+rse/z4cTZu3Mi2bdu48847efXVV/nyyy+9YtIqye6+hyJ4HkYZyOIa0dHR3H777ezdu5f333+fgwcP0tLSwuuvv86vfvUrCgsLEQThrK9RUVHB7bffzltvvcWiRYvk32u1Wq+4Cx588EHef/995s+fzwcffCAHq44cOcIdd9wBQHl5OcuWLSM9PZ2rrrqKBx98UBE8D6L48DzMOTiQxetIKR1PPvkkPT097N27lx07dnDixAmuvvpqcnNzufDCC4f45aqqqvjhD3/IG2+8QXp6+iSuXgEf8uEpgucFzrGBLF6nr6+PoKCgM3ZlJpNJ7ulXUlLCihUryM3NJT4+nptuuonXX3+dCy+8cJJWrTAIRfAUFNxJf38/77//Pm+//Tbvvvsu+/fv5+KLL57sZSkMoAiegoKnsNlsU74t+zmGIngKCgrnDT4jeEqU9jxi//79LFy4kOTkZJ588skzHt+yZQuxsbFkZGSQkZHBq6++OgmrVFDwHOdUx2OF0XE4HGzatGlIiVt2dvYZKRA33HADL7zwwiStUkHBsyg7vPOEzz//nOTkZObOnYtOp+PGG29k165dk70sBQWvogieG6mvrycpKYn29nZgIIs+KSmJmpqayV0YzpW4wUC787S0NNasWTMkYVpB4VxAETw3kpiYyF133SVn1D/44INs2LBhyrR6Wr16NTU1NRQXF3Pttddyyy23TPaSFBTciiJ4bua+++7j8OHDPPvss3zyySeTMT9gRJwpcYuOjpZH+N1xxx0cPXrUq2tUUPA0iuC5GT8/P373u99x33338eyzz/pMPtjy5cupqqqiuroaq9XK1q1byc7OHnKM1NkDBobbpKSkeHuZk05eXh5LlixBrVZz5MiRUY8bK+Kt4JsogucB9u3bx/Tp0yktLZ3spchotVpeeOEFVq5cSUpKCuvWrWPJkiU8+uij8gyF5557jiVLlpCens5zzz3Hli1bJnfRk0Bqaio7duzgiiuuGPUYKeK9b98+ysrK+Mc//kFZWZkXV6ngMqIonu2fwjg5duyYuHjxYrG2tlZMTEwUm5qaJntJCi6wYsUK8YsvvhjxsYMHD4rf/va35Z8ff/xx8fHHH/fW0qYiY+mM1/4pOzw3Iooid911F88++yyzZs3i/vvv9xkfnoL7cDbireB7KILnRv7yl78wa9Ysrr32WgD+8z//k/Lycv79739P8soUBnPNNdeQmpp6xj8lL/HcR6m0cCMbNmxgw4YN8s8ajYYvv/xyElekMBIffPDBhJ7vTMRbwTdRdngKXue2224jLi6O1NTUER8XRZGf/vSnJCcnk5aW5nM3DWci3gq+iSJ4Cl7n1ltvZf/+/aM+vm/fPqqqqqiqquLPf/4zd911l9fWtnPnThISEjh06BDXXXcdK1euBKCpqYlVq1YBo0e8FXwfpT2UwqRQU1NDVlbWiKk7Gzdu5Morr2T9+vXA0IHWClMSn2kPNZbgKSh4BJVKNQfYI4riGXatSqXaAzwpiuInX/18AHhAFMXRM4EVFJxAMWkVFBTOGxTBU/BFGoHBg2MTvvqdgsKEUARPwRfZDdysGuBioEsUxeaxnqSgMBZKHp6C11GpVP8ArgRiVCpVA/BLwA9AFMWXgQJgFXASMAE/mpyVKpxrKEELBQWF8wbFpFVQUDhvUARPQUHhvEERPAUFhfMGRfAUFBTOGxTBU1BQOG9QBE9BQeG8QRE8BQWF84b/D8x9/4WjQ2lQAAAAAElFTkSuQmCC",
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",
       "text/plain": [
        "<Figure size 432x288 with 1 Axes>"
       ]
@@ -157,39 +150,21 @@
     }
    ],
    "source": [
-    "from scipy import special\n",
-    "\n",
-    "def drumhead_height(n, k, distance, angle, t):\n",
-    "   kth_zero = special.jn_zeros(n, k)[-1]\n",
-    "   return np.cos(t) * np.cos(n*angle) * special.jn(n, distance*kth_zero)\n",
-    "\n",
-    "theta = np.r_[0:2*np.pi:50j]\n",
-    "radius = np.r_[0:1:50j]\n",
-    "x = np.array([r * np.cos(theta) for r in radius])\n",
-    "y = np.array([r * np.sin(theta) for r in radius])\n",
-    "z = np.array([drumhead_height(1, 1, r, theta, 0.5) for r in radius])\n",
-    "\n",
-    "import matplotlib.pyplot as plt\n",
-    "fig = plt.figure()\n",
-    "ax = fig.add_axes(rect=(0, 0.05, 0.95, 0.95), projection='3d')\n",
-    "ax.plot_surface(x, y, z, rstride=1, cstride=1, cmap='RdBu_r', vmin=-0.5, vmax=0.5)\n",
-    "ax.set_xlabel('X')\n",
-    "ax.set_ylabel('Y')\n",
-    "ax.set_xticks(np.arange(-1, 1.1, 0.5))\n",
-    "ax.set_yticks(np.arange(-1, 1.1, 0.5))\n",
-    "ax.set_zlabel('Z')\n",
     "\n",
+    "x = np.linspace(0,0.1,2000)\n",
+    "y = np.sin(100 * 2.0*np.pi*x+1.5*np.sin(30 * 2.0*np.pi*x))\n",
+    "plt.plot(x, y, '-')\n",
     "plt.show()"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 18,
+   "execution_count": 12,
    "metadata": {},
    "outputs": [
     {
      "data": {
-      "image/png": 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",
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",
       "text/plain": [
        "<Figure size 432x288 with 1 Axes>"
       ]
@@ -201,11 +176,24 @@
     }
    ],
    "source": [
+    "ratio = 2\n",
+    "first = 1\n",
+    "(length,)  = x.shape\n",
+    "slop = int(length/6)\n",
+    "second = ratio-first\n",
+    "odd = ratio % 2\n",
     "\n",
-    "x = np.linspace(0,0.1,1000)\n",
-    "y = np.sin(100 * 2.0*np.pi*x+1.5*np.sin(30 * 2.0*np.pi*x))\n",
-    "plt.plot(x, y, '-')\n",
-    "plt.show()"
+    "first = int(first * length/ratio) \n",
+    "second = int( second * length/ratio) + odd\n",
+    "slop = np.array(np.append(np.zeros(first-slop) , (np.arange(slop))/slop))\n",
+    "#steep = np.ones(int(first * length/ratio)+ odd) - np.exp(-np.arange(int(first * length/ratio) + odd)/200)\n",
+    "steep = (np.ones(first) + slop)*0.5\n",
+    "\n",
+    "step = np.append(steep,  np.ones(second))\n",
+    "m = np.sin(5 * 2.0 * np.pi * x) * step \n",
+    "plt.plot(x, step, '-')\n",
+    "plt.plot(x, m, '-')\n",
+    "plt.savefig('m_t.pgf', format='pgf')"
    ]
   }
  ],
diff --git a/buch/papers/fm/Python animation/Bessel-FM.py b/buch/papers/fm/Python animation/Bessel-FM.py
index cf30e16..cb35ebd 100644
--- a/buch/papers/fm/Python animation/Bessel-FM.py
+++ b/buch/papers/fm/Python animation/Bessel-FM.py
@@ -4,39 +4,45 @@ from scipy.fft import fft, ifft, fftfreq
 import scipy.special as sc
 import scipy.fftpack
 import matplotlib.pyplot as plt
-from matplotlib.widgets import Slider
-
-# Number of samplepoints
-N = 600
-# sample spacing
-T = 1.0 / 800.0
-x = np.linspace(0.01, N*T, N)
-beta = 1.0
-y_old = np.sin(100.0 * 2.0*np.pi*x+beta*np.sin(50.0 * 2.0*np.pi*x))
-y = 0*x;
-xf = fftfreq(N, 1 / 400)
-for k in range (-5, 5):
-    y = sc.jv(k,beta)*np.sin((100.0+k*50) * 2.0*np.pi*x)
-    yf = fft(y)
-    plt.plot(xf, np.abs(yf))
-
-axbeta =plt.axes([0.25, 0.1, 0.65, 0.03])
-beta_slider = Slider(
-ax=axbeta,
-label="Beta",
-valmin=0.1,
-valmax=3,
-valinit=beta,
-)
-
-def update(val):
-    line.set_ydata(fm(beta_slider.val))
-    fig.canvas.draw_idle()
+import matplotlib as mpl
+# Use the pgf backend (must be set before pyplot imported)
+mpl.use('pgf')
 
+from matplotlib.widgets import Slider
+def fm(beta):
+    # Number of samplepoints
+    N = 600
+    # sample spacing
+    T = 1.0 / 1000.0
+    fc = 100.0
+    fm = 30.0
+    x = np.linspace(0.01, N*T, N)
+    #beta = 1.0
+    y_old = np.sin(fc * 2.0*np.pi*x+beta*np.sin(fm * 2.0*np.pi*x))
+    y = 0*x;
+    xf = fftfreq(N, 1 / N)
+    for k in range (-4, 4):
+        y = sc.jv(k,beta)*np.sin((fc+k*fm) * 2.0*np.pi*x)
+        yf = fft(y)/(fc*np.pi)
+        plt.plot(xf, np.abs(yf))
+    plt.xlim(-150, 150)
+    #plt.savefig('bessel.pgf', format='pgf')
+    plt.show()
 
-beta_slider.on_changed(update)
-plt.show()
+fm(1)
 
-yf_old = fft(y_old)
-plt.plot(xf, np.abs(yf_old))
-plt.show()
\ No newline at end of file
+# Bessel-Funktion
+for n in range (-2,4):
+    x = np.linspace(-11,11,1000)
+    y = sc.jv(n,x)
+    plt.plot(x, y, '-',label='n='+str(n))
+#plt.plot([1,1],[sc.jv(0,1),sc.jv(-1,1)],)
+plt.xlim(-10,10)
+plt.grid(True)
+plt.ylabel('Bessel $J_n(\\beta)$')
+plt.xlabel(' $ \\beta $ ')
+plt.plot(x, y)
+plt.legend()
+#plt.show()
+plt.savefig('bessel.pgf', format='pgf')
+print(sc.jv(0,1))
\ No newline at end of file
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}}%
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+\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}%
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+\pgfpathlineto{\pgfqpoint{0.750000in}{0.500000in}}%
+\pgfpathclose%
+\pgfusepath{fill}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\pgfpathrectangle{\pgfqpoint{0.750000in}{0.500000in}}{\pgfqpoint{4.650000in}{3.020000in}}%
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+\pgfpathlineto{\pgfqpoint{0.750000in}{3.520000in}}%
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+\end{pgfscope}%
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+\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfsetlinewidth{0.803000pt}%
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+\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}%
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+\pgfusepath{stroke,fill}%
+}%
+\begin{pgfscope}%
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+}%
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+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{1.912500in}{0.500000in}%
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+\end{pgfscope}%
+\begin{pgfscope}%
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+}%
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+\end{pgfscope}%
+\end{pgfscope}%
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+\pgfusepath{stroke,fill}%
+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{3.075000in}{0.500000in}%
+\pgfsys@useobject{currentmarker}{}%
+\end{pgfscope}%
+\end{pgfscope}%
+\begin{pgfscope}%
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+}%
+\begin{pgfscope}%
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+\end{pgfscope}%
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+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{4.237500in}{0.500000in}%
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diff --git a/buch/papers/fm/Python animation/m_t.pgf b/buch/papers/fm/Python animation/m_t.pgf
new file mode 100644
index 0000000..edcfb33
--- /dev/null
+++ b/buch/papers/fm/Python animation/m_t.pgf
@@ -0,0 +1,746 @@
+%% Creator: Matplotlib, PGF backend
+%%
+%% To include the figure in your LaTeX document, write
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+%%
+%% 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
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+%%
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diff --git a/buch/papers/fm/Quellen/NaT_Skript_20210920.pdf b/buch/papers/fm/Quellen/NaT_Skript_20210920.pdf
new file mode 100644
index 0000000..b9acc1f
--- /dev/null
+++ b/buch/papers/fm/Quellen/NaT_Skript_20210920.pdf
diff --git a/buch/papers/fm/main.tex b/buch/papers/fm/main.tex
index 731f56f..0c98427 100644
--- a/buch/papers/fm/main.tex
+++ b/buch/papers/fm/main.tex
@@ -14,20 +14,9 @@
 Die Frequenzmodulation ist eine Modulation die man auch schon im alten Radio findet. 
 Falls du dich an die Zeit erinnerst, konnte man zwischen \textit{FM-AM} Umschalten, 
 dies bedeutete so viel wie: \textit{F}requenz-\textit{M}odulation und \textit{A}mplituden-\textit{M}odulation.
-Durch die Modulation wird ein Nachrichtensignal \(m(t)\) auf ein Trägersignal (z.B. ein Sinus- oder Rechtecksignal) abgebildet (kombiniert).
-Durch dieses Auftragen vom Nachrichtensignal \(m(t)\) kann das modulierte Signal in einem gewünschten Frequenzbereich übertragen werden.
-Der ursprünglich Frequenzbereich des Nachrichtensignal \(m(t)\) erstreckt sich typischerweise von 0 Hz bis zur Bandbreite \(B_m\).
-\newline
-Beim Empfänger wird dann durch Demodulation das ursprüngliche Nachrichtensignal \(m(t)\) so originalgetreu wie möglich zurückgewonnen.
-\newline
-Beim Trägersignal \(x_c(t)\) handelt es sich um ein informationsloses Hilfssignal.
-Durch die Modulation mit dem Nachrichtensignal \(m(t)\) wird es zum modulierten zu übertragenden Signal.
-Für alle Erklärungen wird ein sinusförmiges Trägersignal benutzt, jedoch kann auch ein Rechtecksignal,
-welches Digital einfach umzusetzten ist, 
-genauso als Trägersignal genutzt werden kann.
-Zuerst wird erklärt was \textit{FM-AM} ist, danach wie sich diese im Frequenzspektrum verhalten.
-Erst dann erklär ich dir wie die Besselfunktion mit der Frequenzmodulation( acro?) zusammenhängt.
-Nun zur Modulation im nächsten Abschnitt.\cite{fm:NAT}
+Um das Thema einwenig einzuschränken werde ich leider nichts über die Vertiefte, (Physikalische) zusammenhänge oder die Demodulation aufzeigen. 
+Dieses Kapitel soll nurdie Frequenzmodulation und ihren zusammenhang mit der Besselfunktion erklären.
+Aber zuerst einmal zur Modulation selbst, wie funktioniert diese Mathematisch.
 
 
 \input{papers/fm/00_modulation.tex}
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/kra/Makefile.inc b/buch/papers/kra/Makefile.inc
index f453e6e..a521e4b 100644
--- a/buch/papers/kra/Makefile.inc
+++ b/buch/papers/kra/Makefile.inc
@@ -4,11 +4,10 @@
 # (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
 #
 dependencies-kra =						\
-	papers/kra/packages.tex					\
+	papers/kra/packages.tex				\
 	papers/kra/main.tex					\
-	papers/kra/references.bib				\
-	papers/kra/teil0.tex					\
-	papers/kra/teil1.tex					\
-	papers/kra/teil2.tex					\
-	papers/kra/teil3.tex
+	papers/kra/references.bib			\
+	papers/kra/einleitung.tex			\
+	papers/kra/loesung.tex				\
+	papers/kra/anwendung.tex			\
 
diff --git a/buch/papers/kra/anwendung.tex b/buch/papers/kra/anwendung.tex
new file mode 100644
index 0000000..6383984
--- /dev/null
+++ b/buch/papers/kra/anwendung.tex
@@ -0,0 +1,215 @@
+\section{Anwendung \label{kra:section:anwendung}}
+\rhead{Anwendung}
+\newcommand{\dt}[0]{\frac{d}{dt}}
+
+Die Matrix-Riccati Differentialgleichung findet unter anderem Anwendung in der Regelungstechnik beim RQ- und RQG-Regler oder aber auch beim Kalmanfilter.
+Im folgenden Abschnitt möchten wir uns an einem Beispiel anschauen wie wir mit Hilfe der Matrix-Riccati Differentialgleichung (\ref{kra:equation:matrixriccati}) ein Feder-Masse-System untersuchen können \cite{kra:riccati}.
+
+\subsection{Feder-Masse-System}
+Die einfachste Form eines Feder-Masse-Systems ist dargestellt in Abbildung \ref{kra:fig:simple_mass_spring}.
+Es besteht aus einer reibungsfrei gelagerten Masse $m$ ,welche an eine Feder mit der Federkonstante $k$ gekoppelt ist.
+Die im System wirkenden Kräfte teilen sich auf in die auf dem hookeschen Gesetz basierenden Rückstellkraft $F_R = k \Delta_x$ und der auf dem Aktionsprinzip basierenden Kraft $F_a = am = \ddot{x} m$.
+Das Kräftegleichgewicht fordert $F_R = F_a$ woraus folgt, dass
+
+\begin{equation*}
+    k \Delta_x = \ddot{x} m \Leftrightarrow \ddot{x} = \frac{k \Delta_x}{m}
+\end{equation*}
+Die Funktion die diese Differentialgleichung löst, ist die harmonische Schwingung
+\begin{equation}
+    x(t) = A \cos(\omega_0 t + \Phi), \quad \omega_0 = \sqrt{\frac{k}{m}}
+\end{equation}
+\begin{figure}
+    % move image to standalone because the physics package is
+    % incompatible with underbrace
+    \includegraphics{papers/kra/images/simple.pdf}
+    %\input{papers/kra/images/simple_mass_spring.tex}
+    \caption{Einfaches Feder-Masse-System.}
+    \label{kra:fig:simple_mass_spring}
+\end{figure}
+\begin{figure}
+    \input{papers/kra/images/multi_mass_spring.tex}
+    \caption{Feder-Masse-System mit zwei Massen und drei Federn.}
+    \label{kra:fig:multi_mass_spring}
+\end{figure}
+
+\subsection{Hamilton-Funktion}
+Die Bewegung der Masse $m$ kann mit Hilfe der hamiltonschen Mechanik im Phasenraum untersucht werden.
+Die hamiltonschen Gleichungen verwenden dafür die verallgemeinerten Ortskoordinaten
+$q = (q_{1}, q_{2}, ..., q_{n})$ und die verallgemeinerten Impulskoordinaten $p = (p_{1}, p_{2}, ..., p_{n})$, wobei der Impuls definiert ist als $p_k = m_k \cdot v_k$.
+Liegen keine zeitabhängigen Zwangsbedingungen vor, so entspricht die Hamitlon-Funktion der Gesamtenergie des Systems \cite{kra:hamilton}.
+Im Falle des einfachen Feder-Masse-Systems, Abbildung \ref{kra:fig:simple_mass_spring}, setzt sich die Hamilton-Funktion aus kinetischer und potentieller Energie zusammen.
+\begin{equation}
+    \label{kra:harmonischer_oszillator}
+    \begin{split}
+        \mathcal{H}(q, p) &= T(p) + V(q) = E \\
+        &= \underbrace{\frac{p^2}{2m}}_{E_{kin}} + \underbrace{\frac{k q^2}{2}}_{E_{pot}}
+    \end{split}
+\end{equation}
+Die Hamiltonschen Bewegungsgleichungen liefern \cite{kra:kanonischegleichungen}
+\begin{equation}
+    \label{kra:hamilton:bewegungsgleichung}
+    \dot{q_{k}} = \frac{\partial \mathcal{H}}{\partial p_k}
+    \qquad
+    \dot{p_{k}} = -\frac{\partial \mathcal{H}}{\partial q_k}
+\end{equation}
+daraus folgt
+\[
+    \dot{q} = \frac{p}{m}
+    \qquad
+    \dot{p} = -kq
+\]
+in Matrixschreibweise erhalten wir also
+\[
+    \begin{pmatrix}
+        \dot{q} \\
+        \dot{p}
+    \end{pmatrix}
+    =
+    \begin{pmatrix}
+        0  & \frac{1}{m} \\
+        -k & 0
+    \end{pmatrix}
+    \begin{pmatrix}
+        q \\
+        p
+    \end{pmatrix}
+\]
+Für das erweiterte Federmassesystem, Abbildung \ref{kra:fig:multi_mass_spring}, können wir analog vorgehen.
+Die kinetische Energie setzt sich nun aus den kinetischen Energien der einzelnen Massen $m_1$ und $m_2$ zusammen.
+Die Potentielle Energie erhalten wir aus der Summe der kinetischen Energien der einzelnen Federn mit den Federkonstanten $k_1$, $k_c$ und $k_2$.
+\begin{align*}
+    \begin{split}
+        T   &= T_1 + T_2 \\
+        &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2}
+    \end{split}
+    \\
+    \begin{split}
+        V   &= V_1 + V_c + V_2 \\
+        &= \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2}
+    \end{split}
+\end{align*}
+Die Hamilton-Funktion ist also
+\begin{align*}
+    \begin{split}
+        \mathcal{H}     &= T + V \\
+        &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2}
+    \end{split}
+\end{align*}
+Die Bewegungsgleichungen \ref{kra:hamilton:bewegungsgleichung} liefern
+\begin{align*}
+    \frac{\partial \mathcal{H}}{\partial p_k}  & = \dot{q_k}
+    \Rightarrow
+    \left\{
+    \begin{alignedat}{2}
+        \dot{q_1}   &= \frac{2p_1}{2m_1}    &&= \frac{p_1}{m_1}\\
+        \dot{q_2}   &= \frac{2p_2}{2m_2}    &&= \frac{p_2}{m_2}
+    \end{alignedat}
+    \right.
+    \\
+    -\frac{\partial \mathcal{H}}{\partial q_k} & = \dot{p_k}
+    \Rightarrow
+    \left\{
+    \begin{alignedat}{2}
+        \dot{p_1}   &= -(\frac{2k_1q_1}{2} - \frac{2k_c(q_2-q_1)}{2})  &&= -q_1(k_1+k_c) + q_2k_c \\
+        \dot{p_1}   &= -(\frac{2k_c(q_2-q_1)}{2} - \frac{2k_2q_2}{2})  &&= q_1k_c - (k_c + k_2)
+    \end{alignedat}
+    \right.
+\end{align*}
+In Matrixschreibweise erhalten wir
+\begin{equation}
+    \label{kra:hamilton:multispringmass}
+    \begin{pmatrix}
+        \dot{q_1} \\
+        \dot{q_2} \\
+        \dot{p_1} \\
+        \dot{p_2} \\
+    \end{pmatrix}
+    =
+    \begin{pmatrix}
+        0            & 0            & \frac{1}{2m_1} & 0              \\
+        0            & 0            & 0              & \frac{1}{2m_2} \\
+        -(k_1 + k_c) & k_c          & 0              & 0              \\
+        k_c          & -(k_c + k_2) & 0              & 0              \\
+    \end{pmatrix}
+    \begin{pmatrix}
+        q_1 \\
+        q_2 \\
+        p_1 \\
+        p_2 \\
+    \end{pmatrix}
+    \Leftrightarrow
+    \dt
+    \begin{pmatrix}
+        Q \\
+        P \\
+    \end{pmatrix}
+    =
+    \underbrace{
+        \begin{pmatrix}
+            0 & M \\
+            K & 0
+        \end{pmatrix}
+    }_{G}
+    \begin{pmatrix}
+        Q \\
+        P \\
+    \end{pmatrix}
+\end{equation}
+
+\subsection{Phasenraum}
+Der Phasenraum erlaubt die eindeutige Beschreibung aller möglichen Bewegungszustände eines mechanischen Systems durch einen Punkt.
+Die Phasenraumdarstellung eignet sich somit sehr gut für die systematische Untersuchung der Feder-Masse-Systeme.
+
+\subsubsection{Harmonischer Oszillator}
+Die Hamiltonfunktion des harmonischen Oszillators \ref{kra:harmonischer_oszillator} führt auf eine Lösung der Form
+\begin{equation*}
+    q(t) = A \cos(\omega_0 T + \Phi), \quad p(t) = -m \omega_0 A \sin(\omega_0 t + \Phi)
+\end{equation*}
+die Phasenraumtrajektorien bilden also Ellipsen mit Zentrum $q=0, p=0$ und Halbachsen $A$ und $m \omega A$.
+Abbildung \ref{kra:fig:phasenraum} zeigt Phasenraumtrajektorien mit den Energien $E_{x \in \{A, B, C, D\}}$ und verschiedenen Werten von $\omega$.
+\begin{figure}
+    \input{papers/kra/images/phase_space.tex}
+    \caption{Phasenraumdarstellung des einfachen Feder-Masse-Systems.}
+    \label{kra:fig:phasenraum}
+\end{figure}
+
+\subsubsection{Erweitertes Feder-Masse-System}
+Wir intressieren uns nun dafür wie der Phasenwinkel $U = PQ^{-1}$ von der Zeit abhängt,
+wir suchen also die Grösse $\Theta = \dt U$.
+Ersetzten wir in der Gleichung \ref{kra:hamilton:multispringmass} die Matrix $G$ mit $\tilde{G}$ so erhalten wir
+\begin{equation}
+    \dt
+    \begin{pmatrix}
+        Q \\
+        P
+    \end{pmatrix}
+    =
+    \underbrace{
+        \begin{pmatrix}
+            A & B \\
+            C & D
+        \end{pmatrix}
+    }_{\tilde{G}}
+    \begin{pmatrix}
+        Q \\
+        P
+    \end{pmatrix}
+\end{equation}
+Mit einsetzten folgt
+\begin{align*}
+    \dot{Q} = AQ + BP \\
+    \dot{P} = CQ + DP
+\end{align*}
+\begin{equation}
+    \begin{split}
+        \dt U   &= \dot{P} Q^{-1} + P \dt Q^{-1} \\
+        &= (CQ + DP) Q^{-1} - P (Q^{-1} \dot{Q} Q^{-1}) \\
+        &= C\underbrace{QQ^{-1}}_\text{I} + D\underbrace{PQ^{-1}}_\text{U} - P(Q^{-1} (AQ + BP) Q^{-1}) \\
+        &= C + DU - \underbrace{PQ^{-1}}_\text{U}(A\underbrace{QQ^{-1}}_\text{I} + B\underbrace{PQ^{-1}}_\text{U}) \\
+        &= C  + DU - UA - UBU
+    \end{split}
+\end{equation}
+was uns auf die Matrix-Riccati Gleichung \ref{kra:equation:matrixriccati} führt.
+
+% @TODO Einfluss auf anfangsbedingungen, plots?
+% @TODO Fazit ?
diff --git a/buch/papers/kra/einleitung.tex b/buch/papers/kra/einleitung.tex
new file mode 100644
index 0000000..cde2e66
--- /dev/null
+++ b/buch/papers/kra/einleitung.tex
@@ -0,0 +1,14 @@
+\section{Einleitung} \label{kra:section:einleitung}
+\rhead{Einleitung}
+Die riccatische Differentialgleichung ist eine nicht lineare gewöhnliche Differentialgleichung erster Ordnung der Form
+\begin{equation}
+    \label{kra:equation:riccati}
+    y' = f(x)y + g(x)y^2 + h(x)
+\end{equation}
+Sie ist benannt nach dem italienischen Grafen Jacopo Francesco Riccati (1676–1754) der sich mit der Klassifizierung von Differentialgleichungen befasste.
+Als Riccati Gleichung werden auch Matrixgleichungen der Form
+\begin{equation}
+    \label{kra:equation:matrixriccati}
+    \dot{X}(t) = C + DX(t) - X(t)A -X(t)BX(t)
+\end{equation}
+bezeichnet, welche aufgrund ihres quadratischen Terms eine gewisse Ähnlichkeit aufweisen \cite{kra:ethz} \cite{kra:riccati}.
diff --git a/buch/papers/kra/images/Makefile b/buch/papers/kra/images/Makefile
new file mode 100644
index 0000000..ef226a9
--- /dev/null
+++ b/buch/papers/kra/images/Makefile
@@ -0,0 +1,9 @@
+#
+# Makefile -- build standalone images
+#
+# (c) 2022 Prof Dr Andreas Müller
+#
+all:	simple.pdf
+
+simple.pdf:	simple.tex simple_mass_spring.tex
+	pdflatex simple.tex
diff --git a/buch/papers/kra/images/multi_mass_spring.tex b/buch/papers/kra/images/multi_mass_spring.tex
new file mode 100644
index 0000000..f255cc8
--- /dev/null
+++ b/buch/papers/kra/images/multi_mass_spring.tex
@@ -0,0 +1,54 @@
+% create tikz drawing of a multi mass multi spring system
+
+\tikzstyle{vmline}=[red, dashed,line width=0.4,dash pattern=on 1pt off 1pt]
+\tikzstyle{ground}=[pattern=north east lines]
+\tikzstyle{mass}=[line width=0.6,red!30!black,fill=red!40!black!10,rounded corners=1,top color=red!40!black!20,bottom color=red!40!black!10,shading angle=20]
+\tikzstyle{spring}=[line width=0.8,blue!7!black!80,snake=coil,segment amplitude=5,line cap=round]
+
+\begin{tikzpicture}[scale=2]
+    \newcommand{\ticks}[3]
+    {
+        % x, y coordinates
+        \draw[thick] (#1, #2 - 0.1 / 2) --++ (0, 0.1) node[scale=0.8,below=0.2] {#3};
+    }
+    \tikzmath{
+        \hWall = 1.2;
+        \wWall = 0.3;
+        \lWall = 5;
+        \hMass = 0.6;
+        \wMass = 1.1;
+        \xMass1 = 1.0;
+        \xMass2 = 3.0;
+        \xAxisYpos = 0;
+        \originX1 = 0;
+        \originY1 = 0.5;
+        \springscale=7;
+    }
+
+    % create axis
+    \draw[->,thick] (0,\xAxisYpos) --+ (\xMass2 + \wMass, 0) node[right]{$q$};
+    % create ticks on x / q axis
+    \ticks{\xMass1}{\xAxisYpos}{$q_{1}$}
+    \ticks{\xMass2}{\xAxisYpos}{$q_{2}$}
+
+    % create non-moving backgrounds
+    \draw[ground] (\originX1, \originY1) ++ (0, 0) --+(\lWall,0) --+(\lWall, \hWall)
+    --+ (\lWall - \wWall, \hWall) --+(\lWall - \wWall, \wWall) --+ (\wWall, \wWall)  --+(\wWall, \hWall) --+(0, \hWall) -- cycle;
+
+    % create masses
+    \draw[mass] (\originX1, \originY1) ++ (\xMass1, \wWall) rectangle ++ (\wMass,\hMass) node[midway] {$m_{1}$};
+    \draw[mass] (\originX1, \originY1) ++ (\xMass2, \wWall) rectangle ++ (\wMass,\hMass) node[midway] {$m_{2}$};
+
+    % create springs
+    \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++
+    (\wWall, \wWall + \hMass / 2) --++ (\xMass1 - \wWall, 0) node[midway,above=0.2] {$k_1$};
+    \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++
+    (\xMass1 + \wMass, \wWall + \hMass / 2) --++ (\xMass2 - \xMass1 - \wMass, 0) node[midway,above=0.2] {$k_c$};
+    \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++
+    (\xMass2 + \wMass, \wWall + \hMass / 2) --++ (\lWall - \xMass2 - \wMass - \wWall, 0) node[midway,above=0.2] {$k_2$};
+
+    % create vertical measurement line 
+    \draw[vmline] (\xMass1, \xAxisYpos) --+(0, \originY1 + \wWall);
+    \draw[vmline] (\xMass2, \xAxisYpos) --+(0, \originY1 + \wWall);
+
+\end{tikzpicture}
diff --git a/buch/papers/kra/images/phase_space.tex b/buch/papers/kra/images/phase_space.tex
new file mode 100644
index 0000000..cd51ea4
--- /dev/null
+++ b/buch/papers/kra/images/phase_space.tex
@@ -0,0 +1,67 @@
+\colorlet{mypurple}{red!50!blue!90!black!80}
+
+% style to create arrows
+\tikzset{
+    traj/.style 2 args={thick, postaction={decorate},decoration={markings,
+                    mark=at position #1 with {\arrow{<}},
+                    mark=at position #2 with {\arrow{<}}}
+        }
+}
+
+\begin{tikzpicture}[scale=0.6]
+    % p(t=0) = 0, q(t=0) = A, max(p) = mwA
+    \tikzmath{
+        \axh = 5.2;
+        \axw1 = 4.2;
+        \axw2 = 4.8;
+        \d1 = 0.9;
+        \a0 = 1;
+        \b0 = 2;
+        \a1 = \a0 + \d1;
+        \b1 = \b0 + \d1;
+        \a2 = \a1 + \d1;
+        \b2 = \b1 + \d1;
+        \a3 = \a2 + \d1;
+        \b3 = \b2 + \d1;
+        \d2 = 0.75;
+        \aa0 = 2;
+        \bb0 = 1;
+        \aa1 = \aa0 + \d2;
+        \bb1 = \bb0 + \d2;
+        \aa2 = \aa1 + \d2;
+        \bb2 = \bb1 + \d2;
+        \aa3 = \aa2 + \d2;
+        \bb3 = \bb2 + \d2;
+    }
+
+    \draw[->,thick] (-\axw1,0) -- (\axw1,0) node[right] {$q$};
+    \draw[->,thick] (0,-\axh) -- (0,\axh) node[above] {$p$};
+
+    \draw[traj={0.375}{0.875},darkgreen] ellipse (\a0 and \b0);
+    \draw[traj={0.375}{0.875},blue] ellipse (\a1 and \b1);
+    \draw[traj={0.375}{0.875},cyan] ellipse (\a2 and \b2);
+    \draw[traj={0.375}{0.875},mypurple] ellipse (\a3 and \b3);
+
+    \node[right,darkgreen] at (45:{\a0} and {\b0}) {$E_A$};
+    \node[right, blue] at (45:{\a1} and {\b1}) {$E_B$};
+    \node[right, cyan] at (45:{\a2} and {\b2}) {$E_C$};
+    \node[right, mypurple] at (45:{\a3} and {\b3}) {$E_D$};
+    \node[above left] at (110:\b3 + 0.1) {grosses $\omega$};
+
+    \begin{scope}[xshift=12cm]
+        \draw[->,thick] (-\axw2,0) -- (\axw2,0) node[right] {$q$};
+        \draw[->,thick] (0,-\axh) -- (0,\axh) node[above] {$p$};
+
+        \draw[traj={0.375}{0.875},darkgreen] ellipse (\aa0 and \bb0);
+        \draw[traj={0.375}{0.875},blue] ellipse (\aa1 and \bb1);
+        \draw[traj={0.375}{0.875},cyan] ellipse (\aa2 and \bb2);
+        \draw[traj={0.375}{0.875},mypurple] ellipse (\aa3 and \bb3);
+
+        \node[above, darkgreen] at (45:{\aa0} and {\bb0}) {$E_A$};
+        \node[above, blue] at (45:{\aa1} and {\bb1}) {$E_B$};
+        \node[above, cyan] at (45:{\aa2} and {\bb2}) {$E_C$};
+        \node[above, mypurple] at (45:{\aa3} and {\bb3}) {$E_D$};
+
+        \node[above left] at (110:\b3 + 0.1) {kleines $\omega$};
+    \end{scope}
+\end{tikzpicture}
\ No newline at end of file
diff --git a/buch/papers/kra/images/simple.pdf b/buch/papers/kra/images/simple.pdf
new file mode 100644
index 0000000..4351518
--- /dev/null
+++ b/buch/papers/kra/images/simple.pdf
diff --git a/buch/papers/kra/images/simple.tex b/buch/papers/kra/images/simple.tex
new file mode 100644
index 0000000..3bdde27
--- /dev/null
+++ b/buch/papers/kra/images/simple.tex
@@ -0,0 +1,24 @@
+%
+% tikztemplate.tex -- template for standalon tikz images
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\documentclass[tikz]{standalone}
+\usepackage{amsmath}
+\usepackage{times}
+\usepackage{txfonts}
+\usepackage{pgfplots}
+\pgfplotsset{compat=1.16}
+\usepackage[outline]{contour}
+\usepackage{csvsimple}
+\usepackage{physics}
+\usetikzlibrary{arrows,intersections,math}
+\usetikzlibrary{patterns}
+\usetikzlibrary{snakes}
+\usetikzlibrary{arrows.meta}
+\usetikzlibrary{decorations}
+\usetikzlibrary{decorations.markings}
+\begin{document}
+\input{simple_mass_spring.tex}
+\end{document}
+
diff --git a/buch/papers/kra/images/simple_mass_spring.tex b/buch/papers/kra/images/simple_mass_spring.tex
new file mode 100644
index 0000000..868362d
--- /dev/null
+++ b/buch/papers/kra/images/simple_mass_spring.tex
@@ -0,0 +1,66 @@
+% create tikz drawing of a simple mass spring system
+
+\tikzstyle{hmline}=[{Latex[length=3.3,width=2.2]}-{Latex[length=3.3,width=2.2]},line width=0.3]
+\tikzstyle{vmline}=[red, dashed,line width=0.4,dash pattern=on 1pt off 1pt]
+\tikzstyle{ground}=[pattern=north east lines]
+\tikzstyle{mass}=[line width=0.6,red!30!black,fill=red!40!black!10,rounded corners=1,top color=red!40!black!20,bottom color=red!40!black!10,shading angle=20]
+\tikzstyle{spring}=[line width=0.8,blue!7!black!80,snake=coil,segment amplitude=5,line cap=round]
+
+\begin{tikzpicture}[scale=2,>=latex]
+    \newcommand{\ticks}[2]
+    {
+        % arguments: x, y coordinates
+        \draw[thick] (#1, #2 - 0.1 / 2) --++ (0, 0.1);
+    }
+
+    \tikzmath{
+        \hWall = 1.2;
+        \wWall = 0.3;
+        \lWall = 3.5;
+        \hMass = 0.6;
+        \wMass = 1.1;
+        \xMass1 = 1.2;
+        \xMass2 = 2.2;
+        \xAxisYpos = 0;
+        \originX1 = 0;
+        \originY1 = 0.5;
+        \originX2 = 0;
+        \originY2 = -2;
+        \springscale=7;
+    }
+
+    % create x axis
+    \draw[->,thick] (0,\xAxisYpos) --+ (\lWall, 0) node[right]{$x$};
+
+    % create ticks on x axis
+    \ticks{\wWall}{\xAxisYpos}
+    \ticks{\xMass1}{\xAxisYpos}
+    \ticks{\xMass2}{\xAxisYpos}
+
+    % create underground
+    \draw[ground] (\originX1, \originY1) ++ (0, 0) --+(\lWall,0) --+(\lWall, \wWall) --+(\wWall, \wWall) --+(\wWall, \hWall) --+(0, \hWall) -- cycle;
+    \draw[ground] (\originX2, \originY2) ++ (0, 0) --+(\lWall,0) --+(\lWall, \wWall) --+(\wWall, \wWall) --+(\wWall, \hWall) --+(0, \hWall) -- cycle;
+
+    % create masses
+    \draw[mass] (\originX1, \originY1) ++ (\xMass1, \wWall) rectangle ++ (\wMass,\hMass) node[midway] {$m$};
+    \draw[mass] (\originX2, \originY2) ++ (\xMass2, \wWall) rectangle ++ (\wMass,\hMass) node[midway] {$m$};
+
+    % create springs
+    \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++
+    (\wWall, \wWall + \hMass / 2) --++ (\xMass1 - \wWall, 0) node[midway,above=3.5] {$k$};
+    \draw[spring, segment length=(\xMass2 - \wWall) * \springscale] (\originX2, \originY2) ++
+    (\wWall, \wWall + \hMass / 2) --++ (\xMass2 - \wWall, 0) node[midway,above=3.5] {$k$};
+
+    % create vertical measurement line 
+    \draw[vmline] (\xMass1, \xAxisYpos) --+(0, \originY1 + \wWall);
+    \draw[vmline] (\xMass2, \xAxisYpos) --+(0, \originY2 + \hMass+\wWall);
+    \draw[vmline] (\wWall, \originY1+\wWall) --(\wWall, \originY2 + \hWall);
+
+    % create horizontal measurement line
+    \draw[hmline] (\wWall, \xAxisYpos + 0.2) -- (\xMass1, \xAxisYpos + 0.2) node[midway,fill=white,inner sep=0] {$l_0$};
+    \draw[hmline] (\xMass1, \xAxisYpos + 0.2) -- (\xMass2, \xAxisYpos + 0.2) node[midway,fill=white,inner sep=0] {$\Delta_{x}$};
+    \draw[hmline] (\wWall, \xAxisYpos - 0.3) -- (\xMass2, \xAxisYpos - 0.3) node[midway,fill=white,inner sep=0] {$l_{1}$};
+
+    %  create force arrow
+    \draw[->,blue, very thick,line cap=round] (\xMass2 + \wMass / 2, \originY2 + \wWall + \hMass + 0.15) node[above] {$\vb{F_{R}}$} --+ (-0.5, 0);
+\end{tikzpicture}
diff --git a/buch/papers/kra/loesung.tex b/buch/papers/kra/loesung.tex
new file mode 100644
index 0000000..4e0da1c
--- /dev/null
+++ b/buch/papers/kra/loesung.tex
@@ -0,0 +1,86 @@
+\section{Lösungsmethoden} \label{kra:section:loesung}
+\rhead{Lösungsmethoden}
+
+\subsection{Riccatische Differentialgleichung} \label{kra:loesung:riccati}
+Eine allgemeine analytische Lösung der Riccati Differentialgleichung ist nicht möglich.
+Es gibt aber Spezialfälle, in denen sich die Gleichung vereinfachen lässt und so eine analytische Lösung gefunden werden kann.
+Diese wollen wir im folgenden Abschnitt genauer anschauen.
+
+\subsubsection{Fall 1: Konstante Koeffizienten}
+Sind die Koeffizienten $f(x), g(x), h(x)$ Konstanten, so lässt sich die DGL separieren und reduziert sich auf die Lösung des Integrals \ref{kra:equation:case1_int}.
+\begin{equation}
+    y' = fy^2 + gy + h 
+\end{equation}
+\begin{equation}
+    \frac{dy}{dx} = fy^2 + gy + h
+\end{equation}
+\begin{equation} \label{kra:equation:case1_int}
+    \int \frac{dy}{fy^2 + gy + h} = \int dx
+\end{equation}
+
+\subsubsection{Fall 2: Bekannte spezielle Lösung}
+Kennt man eine spezielle Lösung $y_p$ so kann die riccatische DGL mit Hilfe einer Substitution auf eine lineare Gleichung reduziert werden.
+Wir wählen als Substitution
+\begin{equation} \label{kra:equation:substitution}
+    z = \frac{1}{y - y_p} 
+\end{equation}
+durch Umstellen von \ref{kra:equation:substitution} folgt
+\begin{equation}
+    y = y_p + \frac{1}{z^2} \label{kra:equation:backsubstitution}
+\end{equation}
+\begin{equation}
+    y' = y_p' - \frac{1}{z^2}z'
+\end{equation}
+mit Einsetzten in die DGL \ref{kra:equation:riccati} folgt 
+\begin{equation}
+    y_p' - \frac{1}{z^2}z' = f(x)(y_p + \frac{1}{z}) + g(x)(y_p + \frac{1}{z})^2 + h(x)
+\end{equation}
+\begin{equation}
+    -z^{2}y_p' + z' = -z^2\underbrace{(y_{p}f(x) + g(x)y_p^2 + h(x))}_{y_p'} - z(f(x) + 2y_{p}g(x)) - g(x)
+\end{equation}
+was uns direkt auf eine lineare Differentialgleichung 1.Ordnung führt.
+\begin{equation}
+    z' = -z(f(x) + 2y_{p}g(x)) - g(x)
+\end{equation}
+Diese kann nun mit den Methoden zur Lösung von linearen Differentialgleichungen 1.Ordnung gelöst werden.
+Durch die Rücksubstitution \ref{kra:equation:backsubstitution} erhält man dann die Lösung von \ref{kra:equation:riccati}.
+
+\subsection{Matrix-Riccati Differentialgleichung} \label{kra:loesung:riccati}
+% Lösung matrix riccati
+Die Lösung der Matrix-Riccati Gleichung \ref{kra:equation:matrixriccati} erhalten wir nach \cite{kra:kalmanisae} folgendermassen
+\begin{equation}
+    \label{kra:matrixriccati-solution}
+    \begin{pmatrix}
+        X(t) \\
+        Y(t)
+    \end{pmatrix}
+    =
+    \Phi(t_0, t)
+    \begin{pmatrix}
+        I(t) \\
+        U_0(t)
+    \end{pmatrix}
+    =
+    \begin{pmatrix}
+        \Phi_{11}(t_0, t) & \Phi_{12}(t_0, t) \\
+        \Phi_{21}(t_0, t) & \Phi_{22}(t_0, t)
+    \end{pmatrix}
+    \begin{pmatrix}
+        I(t) \\
+        U_0(t)
+    \end{pmatrix}
+\end{equation}
+\begin{equation}
+    U(t) =
+    \begin{pmatrix}
+        \Phi_{21}(t_0, t) + \Phi_{22}(t_0, t)
+    \end{pmatrix}
+    \begin{pmatrix}
+        \Phi_{11}(t_0, t) + \Phi_{12}(t_0, t)
+    \end{pmatrix}
+    ^{-1}
+\end{equation}
+wobei $\Phi(t, t_0)$ die sogenannte Zustandsübergangsmatrix ist.
+\begin{equation}
+    \Phi(t_0, t) = e^{H(t - t_0)}
+\end{equation}
diff --git a/buch/papers/kra/main.tex b/buch/papers/kra/main.tex
index fcee25b..a84ebaf 100644
--- a/buch/papers/kra/main.tex
+++ b/buch/papers/kra/main.tex
@@ -3,34 +3,12 @@
 %
 % (c) 2020 Hochschule Rapperswil
 %
-\chapter{Kalman, Riccati und Abel\label{chapter:kra}}
-\lhead{Kalman, Riccati und Abel}
+\chapter{Riccati Differentialgleichung\label{chapter:kra}}
+\lhead{Riccati Differentialgleichung}
 \begin{refsection}
-    \chapterauthor{Samuel Niederer}
-
-    Ein paar Hinweise für die korrekte Formatierung des Textes
-    \begin{itemize}
-        \item
-              Absätze werden gebildet, indem man eine Leerzeile einfügt.
-              Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet.
-        \item
-              Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende
-              Optionen werden gelöscht.
-              Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen.
-        \item
-              Beginnen Sie jeden Satz auf einer neuen Zeile.
-              Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen
-              in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt
-              anzuwenden.
-        \item
-              Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren
-              Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern.
-    \end{itemize}
-
-    \input{papers/kra/teil0.tex}
-    \input{papers/kra/teil1.tex}
-    \input{papers/kra/teil2.tex}
-    \input{papers/kra/teil3.tex}
-
-    \printbibliography[heading=subbibliography]
+      \chapterauthor{Samuel Niederer}
+      \input{papers/kra/einleitung.tex}
+      \input{papers/kra/loesung.tex}
+      \input{papers/kra/anwendung.tex}
+      \printbibliography[heading=subbibliography]
 \end{refsection}
diff --git a/buch/papers/kra/packages.tex b/buch/papers/kra/packages.tex
index df34dcf..56c48d9 100644
--- a/buch/papers/kra/packages.tex
+++ b/buch/papers/kra/packages.tex
@@ -8,3 +8,11 @@
 % following example
 %\usepackage{packagename}
 
+%\usepackage{physics}
+\usepackage[outline]{contour}
+\pgfplotsset{compat=1.16}
+\usetikzlibrary{patterns}
+\usetikzlibrary{snakes}
+\usetikzlibrary{arrows.meta}
+\usetikzlibrary{decorations}
+\usetikzlibrary{decorations.markings}
diff --git a/buch/papers/kra/presentation/presentation.tex b/buch/papers/kra/presentation/presentation.tex
new file mode 100644
index 0000000..eb6541b
--- /dev/null
+++ b/buch/papers/kra/presentation/presentation.tex
@@ -0,0 +1,491 @@
+\documentclass[ngerman, aspectratio=169, xcolor={rgb}]{beamer}
+
+% style
+\mode<presentation>{
+	\usetheme{Frankfurt}
+}
+%packages
+\usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-}
+\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{booktabs} %toprule midrule bottomrue in tables
+\usepackage{scrextend}
+\usepackage{textgreek}
+\usepackage[rgb]{xcolor}
+
+\usepackage[normalem]{ulem} % \sout
+
+\usepackage{ marvosym } % \Lightning
+
+\usepackage{multimedia} % embedded videos
+
+\usepackage{tikz}
+\usepackage{pgf}
+\usepackage{pgfplots}
+
+\usepackage{algorithmic}
+
+%citations
+\usepackage[style=verbose,backend=biber]{biblatex}
+\addbibresource{references.bib}
+
+
+%math font
+\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{KRA}
+\subtitle{Kalman Riccati Abel}
+\author{Samuel Niederer}
+% \institute{OST Ostschweizer Fachhochschule}
+% \institute{\includegraphics[scale=0.3]{../img/ost_logo.png}}
+\date{\today}
+
+\input{../packages.tex}
+
+\newcommand*{\QED}{\hfill\ensuremath{\blacksquare}}%
+
+\newcommand*{\HL}{\textcolor{mainColor}}
+\newcommand*{\RD}{\textcolor{red}}
+\newcommand*{\BL}{\textcolor{blue}}
+\newcommand*{\GN}{\textcolor{dgreen}}
+\newcommand{\dt}[0]{\frac{d}{dt}}
+
+\definecolor{darkgreen}{rgb}{0,0.6,0}
+
+
+\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,shapes.geometric, fadings}
+
+\begin{document}
+
+\begin{frame}
+	\titlepage
+\end{frame}
+
+\begin{frame}
+	\frametitle{Content}
+	\tableofcontents
+\end{frame}
+
+\section{Einführung}
+
+\begin{frame}
+	\begin{itemize}
+		\item<1|only@1> \textbf{K}alman
+		\item<1|only@1> \textbf{R}iccati
+		\item<1|only@1> \textbf{A}bel
+
+		\item<2|only@2> \textcolor{red}{\sout{\textbf{K}alman}}
+		\item<2|only@2> \textbf{R}iccati
+		\item<2|only@2> \textbf{A}bel
+
+		\item<3|only@3> \textcolor{red}{\sout{\textbf{K}alman}} \textcolor{green}{Federmassesytem}
+		\item<3|only@3> \textbf{R}iccati
+		\item<3|only@3> \textbf{A}bel
+
+		\item<4|only@4> \textcolor{red}{\sout{\textbf{K}alman}} \textcolor{green}{Federmassesytem}
+		\item<4|only@4> \textbf{R}iccati
+		\item<4|only@4> \uwave{\textbf{A}bel}
+	\end{itemize}
+\end{frame}
+
+\section{Riccati}
+
+\begin{frame}
+	\frametitle{Riccatische Differentialgleichung}
+	\begin{equation*}
+		% y'(x) = f(x)y^2(x) + g(x)y(x) + h(x)
+		x'(t) = f(t)x^2(t) + g(t)x(t) + h(t)
+	\end{equation*}
+
+	\pause
+
+	\begin{equation*}
+		\dot{X}(t) = - X(t)BX(t) - X(t)A + DX(t) + C
+	\end{equation*}
+
+	% \pause
+	% Anwendungen
+	% \begin{itemize}
+	% 	\item Zeitkontinuierlicher Kalmanfilter
+	% 	\item Regelungstechnik LQ-Regler
+	% 	\item Federmassesyteme
+	% \end{itemize}
+\end{frame}
+
+\begin{frame}
+	\frametitle{Auftreten der Gleichung}
+	\begin{columns}
+		\column{0.4 \textwidth}
+		\begin{equation*}
+			\dt
+			\begin{pmatrix}
+				X \\
+				Y
+			\end{pmatrix}
+			=
+			\underbrace{
+				\begin{pmatrix}
+					A & B \\
+					C & D
+				\end{pmatrix}
+			}_{H}
+			\begin{pmatrix}
+				X \\
+				Y
+			\end{pmatrix}
+		\end{equation*}
+
+		\pause
+
+		\column{0.4 \textwidth}
+		\begin{equation*}
+			U = YX^{-1} \qquad \dt U = ?
+		\end{equation*}
+	\end{columns}
+
+	\pause
+
+	\begin{align*}
+		\dt U         & = \dot{Y} X^{-1} + Y \dt X^{-1}                                                                                                    \\
+		\uncover<4->{ & = (CX + DY) X^{-1} - Y (X^{-1} \dot{X} X^{-1})\\}
+		\uncover<5->{ & = C\underbrace{XX^{-1}}_\text{I} + D\underbrace{YX^{-1}}_\text{U} - Y(X^{-1} (AX + BY) X^{-1})\\}
+		\uncover<6->{ & = C + DU - \underbrace{YX^{-1}}_\text{U}(A\underbrace{XX^{-1}}_\text{I} + B\underbrace{YX^{-1}}_\text{U})\\}
+		\uncover<7->{ & = C  + DU - UA - UBU}
+	\end{align*}
+\end{frame}
+
+\begin{frame}
+	\frametitle{Lösen der Gleichung}
+	\begin{equation*}
+		\begin{pmatrix}
+			X(t) \\
+			Y(t)
+		\end{pmatrix}
+		=
+		\Phi(t_0, t)
+		\begin{pmatrix}
+			I(t) \\
+			U_0(t)
+		\end{pmatrix}
+		=
+		\begin{pmatrix}
+			\Phi_{11}(t_0, t) & \Phi_{12}(t_0, t) \\
+			\Phi_{21}(t_0, t) & \Phi_{22}(t_0, t)
+		\end{pmatrix}
+		\begin{pmatrix}
+			I(t) \\
+			U_0(t)
+		\end{pmatrix}
+	\end{equation*}
+
+	\pause
+
+	\begin{equation*}
+		U(t) =
+		\begin{pmatrix}
+			\Phi_{21}(t_0, t) + \Phi_{22}(t_0, t) U_0(t)
+		\end{pmatrix}
+		\begin{pmatrix}
+			\Phi_{11}(t_0, t) + \Phi_{12}(t_0, t) U_0(t)
+		\end{pmatrix}
+		^{-1}
+	\end{equation*}
+
+	\pause
+
+	% wobei $\Phi(t, t_0)$ die sogennante Zustandsübergangsmatrix ist.
+
+	\begin{equation*}
+		\Phi(t_0, t) = e^{H(t - t_0)}
+	\end{equation*}
+\end{frame}
+
+\section{Federmassystem}
+\begin{frame}
+	\frametitle{Federmassesystem}
+	\begin{columns}
+		\column{0.5 \textwidth}
+		\input{../images/simple_mass_spring.tex}
+
+		\column{0.5 \textwidth}
+		\begin{align*}
+			\uncover<2->{F_R      & = k \Delta_x  \\}
+			\uncover<3->{F_a      & = am = \ddot{x} m \\}
+			\uncover<4->{F_R      & = F_a \Leftrightarrow k \Delta_x = \ddot{x} m\\}
+			\uncover<5->{\ddot{x} & = \frac{k \Delta_x}{m} \\}
+			\uncover<6->{x(t)     & = A \cos(\omega_0 + \Phi), \quad \omega_0 = \sqrt{\frac{k}{m}}}
+		\end{align*}
+	\end{columns}
+\end{frame}
+
+\begin{frame}
+	\frametitle{Phasenraum}
+	\begin{columns}
+		\column{0.3 \textwidth}
+		\begin{tikzpicture}[scale=3]
+			\draw[->, thick] (0, 0) -- (1,0) node[right] {$q$};
+			\draw[->, thick] (0.5, -0.5) -- (0.5,0.5) node[above]{$p$};
+		\end{tikzpicture}
+		\column{0.7 \textwidth}
+		Impulskoordinaten $p = (p_{1}, p_{2}, ..., p_{n}), \quad p=mv$ \\
+		Ortskoordinaten $q = (q_{1}, q_{2}, ..., q_{n})$ \\
+
+
+
+		\begin{align*}
+			\uncover<2->{\mathcal{H}(q, p) & = \underbrace{T(p)}_{E_{kin}} + \underbrace{V(q)}_{E_{pot}} = E_{tot} \\}
+			\uncover<3->{                  & = \frac{p^2}{2m}+ \frac{k q^2}{2}}
+		\end{align*}
+
+
+
+		\begin{equation*}
+			\uncover<4->{
+				\dot{q_{k}} = \frac{\partial \mathcal{H}}{\partial p_k}
+				\qquad
+				\dot{p_{k}} = -\frac{\partial \mathcal{H}}{\partial q_k}
+			}
+		\end{equation*}
+
+		\pause
+
+		\begin{equation*}
+			\uncover<5->{
+				\begin{pmatrix}
+					\dot{q} \\
+					\dot{p}
+				\end{pmatrix}
+				=
+				\begin{pmatrix}
+					0  & \frac{1}{m} \\
+					-k & 0
+				\end{pmatrix}
+				\begin{pmatrix}
+					q \\
+					p
+				\end{pmatrix}
+			}
+		\end{equation*}
+
+	\end{columns}
+\end{frame}
+
+\begin{frame}
+	\frametitle{Phasenraum}
+	\input{../images/phase_space.tex}
+\end{frame}
+
+\begin{frame}
+	\frametitle{Federmassesystem}
+	\begin{columns}
+		\column{0.6 \textwidth}
+		\scalebox{0.8}{\input{../images/multi_mass_spring.tex}}
+		\begin{align*}
+			\uncover<2->{\mathcal{H} & = T + V \\}
+			\uncover<7->{            & = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2}}
+		\end{align*}
+
+		\column{0.4 \textwidth}
+		\begin{align*}
+			\uncover<3->{T & = T_1 + T_2}                                                               \\
+			\uncover<5->{  & = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2}   }                              \\
+			\uncover<4->{V & = V_1 + V_c + V_2 }                                                        \\
+			\uncover<6->{  & = \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2}}
+		\end{align*}
+	\end{columns}
+\end{frame}
+
+\begin{frame}
+	\frametitle{Federmassesystem}
+	\begin{equation*}
+		\begin{pmatrix}
+			\dot{q_1} \\
+			\dot{q_2} \\
+			\dot{p_1} \\
+			\dot{p_2} \\
+		\end{pmatrix}
+		=
+		\begin{pmatrix}
+			0            & 0            & \frac{1}{2m_1} & 0              \\
+			0            & 0            & 0              & \frac{1}{2m_2} \\
+			-(k_1 + k_c) & k_c          & 0              & 0              \\
+			k_c          & -(k_c + k_2) & 0              & 0              \\
+		\end{pmatrix}
+		\begin{pmatrix}
+			q_1 \\
+			q_2 \\
+			p_1 \\
+			p_2 \\
+		\end{pmatrix}
+		\Leftrightarrow
+		\dt
+		\begin{pmatrix}
+			Q \\
+			P \\
+		\end{pmatrix}
+		\underbrace{
+			\begin{pmatrix}
+				0 & M \\
+				K & 0
+			\end{pmatrix}
+		}_{H}
+		\begin{pmatrix}
+			Q \\
+			P \\
+		\end{pmatrix}
+	\end{equation*}
+
+	\pause
+
+	$U = PQ^{-1} \qquad \dt U = ?$
+
+	\pause
+
+	\begin{align*}
+		\dt U & = C  + DU - UA - UBU \\
+		      & = K - UMU
+	\end{align*}
+
+\end{frame}
+
+\begin{frame}
+	\frametitle{Einfluss der Anfangsbedingung:}
+	\begin{columns}
+		\column{0.4 \textwidth}
+		\begin{equation*}
+			\uncover<2->{q_0 =
+				\begin{pmatrix}
+					q_{10} \\
+					q_{20}
+				\end{pmatrix}
+				=
+				\begin{pmatrix}
+					3 \\
+					1
+				\end{pmatrix}
+			}
+		\end{equation*}
+		\begin{equation*}
+			\uncover<3->{q_0 =
+				\begin{pmatrix}
+					q_{10} \\
+					q_{20}
+				\end{pmatrix}
+				=
+				\begin{pmatrix}
+					3 \\
+					3
+				\end{pmatrix}
+			}
+		\end{equation*}
+		\begin{equation*}
+			\uncover<4->{q_0 =
+				\begin{pmatrix}
+					q_{10} \\
+					q_{20}
+				\end{pmatrix}
+				=
+				\begin{pmatrix}
+					2 \\
+					-2
+				\end{pmatrix}
+			}
+		\end{equation*}
+		\column{0.6 \textwidth}
+		\scalebox{0.8}{\input{../images/multi_mass_spring.tex}}
+	\end{columns}
+\end{frame}
+
+\section{Schlussteil}
+\begin{frame}
+	\frametitle{Zusammenfassung}
+	\begin{itemize}
+		\pause
+		\item{Riccatische Differentialgleichung}
+		      \pause
+		      \begin{itemize}
+			      \item{Ausgansgleichung}
+			            \pause
+			      \item{Lösung}
+		      \end{itemize}
+		      \pause
+		\item{Harmonischer Ozillator}
+		      \pause
+		      \begin{itemize}
+			      \item{Hamiltonfunktion}
+			            \pause
+			      \item{Phasenraum}
+		      \end{itemize}
+		      \pause
+		\item{Gekoppelter harmonischer Ozillator}
+		      \pause
+		      \begin{itemize}
+			      \item{Riccatische Differentialgleichung}
+			            \pause
+			      \item{Einfluss der Anfangsbedingungen}
+		      \end{itemize}
+		      \pause
+		\item{\uwave{Abel}}
+		      \begin{itemize}
+			      \pause
+			      \item{Nichtlineare Federkonstante}
+		      \end{itemize}
+
+	\end{itemize}
+\end{frame}
+
+\end{document}
diff --git a/buch/papers/kra/references.bib b/buch/papers/kra/references.bib
index f13c3d8..a9a8ede 100644
--- a/buch/papers/kra/references.bib
+++ b/buch/papers/kra/references.bib
@@ -4,32 +4,42 @@
 % (c) 2020 Autor, Hochschule Rapperswil
 %
 
-@online{kra:bibtex,
-	title = {BibTeX},
-	url = {https://de.wikipedia.org/wiki/BibTeX},
-	date = {2020-02-06},
-	year = {2020},
-	month = {2},
-	day = {6}
+@misc{kra:riccati,
+title = {Riccatische Differentialgleichung},
+url = {https://de.wikipedia.org/wiki/Riccatische_Differentialgleichung},
+date   = {2022-05-26}
 }
 
-@book{kra: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}
+@misc{kra:ethz,
+author = {Ch. Roduner},
+title = {Die-Riccati-Gleichung},
+url = {https://www.imrtweb.ethz.ch/users/geering/Riccati.pdf},
+date   = {2022-05-26}
 }
 
-@article{kra: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{kra:hamilton,
+  title = {Hamilton-Funktion},
+  url   = {https://de.wikipedia.org/wiki/Hamilton-Funktion},
+  date  = {2022-05-26}
 }
 
+@misc{kra:kanonischegleichungen,
+  title = {Kanonische Gleichungen},
+  url   = {https://de.wikipedia.org/wiki/Kanonische_Gleichungen},
+  date  = {2022-05-26}
+}
+
+@misc{kra:newton,
+  title = {Newtonsche Gesetze},
+  url   = {https://de.wikipedia.org/wiki/Newtonsche_Gesetze},
+  date  = {2022-05-26}
+}
+
+@misc{kra:kalmanisae,
+  author = {D.Alazard},
+  title  = {Introduction to Kalman filtering},
+  url    = {https://pagespro.isae-supaero.fr/IMG/pdf/introKalman_e_151211.pdf},
+  date   = {2022-05-26}
+}
+
+
diff --git a/buch/papers/kra/scripts/animation.py b/buch/papers/kra/scripts/animation.py
new file mode 100644
index 0000000..5e805ae
--- /dev/null
+++ b/buch/papers/kra/scripts/animation.py
@@ -0,0 +1,243 @@
+import numpy as np

+import matplotlib.pyplot as plt

+import matplotlib.patches

+import matplotlib.transforms

+import matplotlib.text

+from matplotlib.animation import FuncAnimation

+import imageio

+

+from simulation import Simulation

+

+

+class Mass:

+    def __init__(self, x_0, width, height, **kwargs):

+        self._x_0 = x_0

+        xy = (x_0, 0)

+        self._rect = matplotlib.patches.Rectangle(xy, width, height, **kwargs)

+

+    @property

+    def patch(self):

+        return self._rect

+

+    @property

+    def x(self):

+        return self._rect.get_x()

+

+    @property

+    def width(self):

+        return self._rect.get_width()

+

+    def move(self, x):

+        self._rect.set_x(self._x_0 + x)

+

+

+class Spring:

+    def __init__(self, n, height, ax, resolution=1000, **kwargs):

+        self._n = n

+        self._height = height

+        self._N = resolution

+        (self._line,) = ax.plot([], [], "-", **kwargs)

+

+    def set(self, x_0, x_1):

+        T = (x_1 - x_0) / self._n

+        x = np.linspace(x_0, x_1, self._N, endpoint=True)

+        t = np.linspace(0, x_1 - x_0, self._N)

+        y = (np.sin(2 * np.pi * t / T) + 1.5) * self._height / 2

+        self.line.set_data(x, y)

+

+    @property

+    def line(self):

+        return self._line

+

+

+class LinePlot:

+    def __init__(self, ax, **kwargs):

+        (self._line,) = ax.plot([], [], "-", **kwargs)

+        self._x = []

+        self._y = []

+

+    @property

+    def line(self):

+        return self._line

+

+    def update(self, x, y):

+        self._x.append(x)

+        self._y.append(y)

+        self._line.set_data(self._x, self._y)

+

+

+class ScatterPlot:

+    def __init__(self, ax, **kwargs):

+        self._color = kwargs.get("color", "tab:green")

+        self._line = ax.scatter([], [], **kwargs)

+        self._ax = ax

+        self._x = []

+        self._y = []

+

+    @property

+    def line(self):

+        return self._line

+

+    def update(self, x, y, **kwargs):

+        self._x.append(x)

+        self._y.append(y)

+        self._line.remove()

+        self._line = self._ax.scatter(self._x, self._y, color=self._color, **kwargs)

+

+

+class QuiverPlot:

+    def __init__(self, ax, **kwargs):

+        self.x = []

+        self.y = []

+        self.u = []

+        self.v = []

+        self.ax = ax

+        self.ln = self.ax.quiver([], [], [], [])

+

+    def update(self, x, y, u, v):

+        self.x.append(x)

+        self.y.append(y)

+        self.u.append(u)

+        self.v.append(v)

+        self.ln.remove()

+        self.ln = self.ax.quiver(self.x, self.y, self.u, self.v)

+

+    @property

+    def line(self):

+        return self.ln

+

+

+anim_folder = "anim_0"

+img_counter = 0

+

+sim = Simulation()

+params = {

+    "x_0": [2, -2],

+    "k_1": 1,

+    "k_c": 2,

+    "k_2": 1,

+    "m_1": 0.5,

+    "m_2": 0.5,

+}

+

+time = 2.1

+

+

+# create axis

+fig = plt.figure(figsize=(20, 15), constrained_layout=True)

+fig.suptitle(

+    " ,".join([f"${key} = {val}$" for (key, val) in params.items()]), fontsize=20

+)

+spec = fig.add_gridspec(3, 4)

+ax0 = fig.add_subplot(spec[-1, :])

+ax1 = fig.add_subplot(spec[:-1, :2])

+ax2 = fig.add_subplot(spec[:-1, 2:])

+

+ax0.set_yticks([])

+

+mass_height = 0.5

+spring_height = 0.6 * mass_height

+x_max = 21

+y_max = 2 * mass_height

+

+mass_1 = Mass(

+    7,

+    2,

+    mass_height,

+    color="tab:red",

+)

+mass_2 = Mass(14, 2, mass_height, color="tab:blue")

+masses = [mass_1, mass_2]

+patches = [mass.patch for mass in masses]

+

+spring_1 = Spring(4, spring_height, ax0, color="tab:red", linewidth=10)

+spring_2 = Spring(4, spring_height, ax0, color="tab:gray", linewidth=10)

+spring_3 = Spring(4, spring_height, ax0, color="tab:blue", linewidth=10)

+springs = [spring_1, spring_2, spring_3]

+

+linePlot_1 = LinePlot(ax1, color="tab:red", label="$m_1$", alpha=1)

+linePlot_2 = LinePlot(ax1, color="tab:blue", label="$m_2$", alpha=1)

+linePlots = [linePlot_1, linePlot_2]

+

+# quiverPlot = QuiverPlot(ax2)

+scatterPlot = ScatterPlot(ax2)

+

+lines = [spring.line for spring in springs]

+lines.extend([plot.line for plot in linePlots])

+# lines.append(quiverPlot.line)

+lines.append(scatterPlot.line)

+

+objects = lines + patches

+

+ax0.plot(

+    np.repeat(mass_1.x, 2),

+    [0, y_max],

+    "--",

+    color="tab:red",

+    label="Ruhezustand $m_1$",

+)

+ax0.plot(

+    np.repeat(mass_2.x, 2),

+    [0, y_max],

+    "--",

+    color="tab:blue",

+    label="Ruhezustand $m_2$",

+)

+

+

+def init():

+    ax0.set_xlim(0, x_max)

+    ax0.set_ylim(0, y_max)

+

+    ax1.set_xlim(0, time)

+    ax1.set_ylim(-4, 4)

+    ax1.set_xlabel("time", fontsize=20)

+    ax1.set_ylabel("$q$", fontsize=20)

+

+    ax2.set_xlim(-4, 4)

+    ax2.set_ylim(-4, 4)

+    ax2.set_xlabel("$q_1$", fontsize=20)

+    ax2.set_ylabel("$q_2$", fontsize=20)

+

+    for patch in patches:

+        ax0.add_patch(patch)

+

+    spring_1.set(0, mass_1.x)

+    spring_2.set(mass_1.x + mass_1.width, mass_2.x)

+    spring_2.set(mass_2.x + mass_2.width, x_max)

+

+    return objects

+

+

+def update(frame):

+    global img_counter

+    x_1, x_2 = sim(frame, **params)

+

+    mass_1.move(x_1)

+    mass_2.move(x_2)

+

+    spring_1.set(0, mass_1.x)

+    spring_2.set(mass_1.x + mass_1.width, mass_2.x)

+    spring_3.set(mass_2.x + mass_2.width, x_max)

+

+    linePlot_1.update(frame, x_1)

+    linePlot_2.update(frame, x_2)

+

+    scatterPlot.update(x_1, x_2, alpha=0.25)

+

+    img_counter += 1

+    return objects

+

+

+anim = FuncAnimation(

+    fig,

+    update,

+    frames=np.linspace(0, time, int(time * 30)),

+    init_func=init,

+    blit=False,

+)

+

+ax0.legend(fontsize=20)

+ax1.legend(fontsize=20)

+FFwriter = matplotlib.animation.FFMpegWriter(fps=30)

+anim.save("animation.mp4", writer=FFwriter)

diff --git a/buch/papers/kra/scripts/simulation.py b/buch/papers/kra/scripts/simulation.py
new file mode 100644
index 0000000..8bccb6a
--- /dev/null
+++ b/buch/papers/kra/scripts/simulation.py
@@ -0,0 +1,40 @@
+import sympy as sp

+

+

+class Simulation:

+    def __init__(self):

+        self.k_1, self.k_2, self.k_c = sp.symbols("k_1 k_2 k_c")

+        self.m_1, self.m_2 = sp.symbols("m_1 m_2")

+        self.t = sp.symbols("t")

+        K = sp.Matrix(

+            [[-(self.k_1 + self.k_c), self.k_c], [self.k_c, -(self.k_2 + self.k_c)]]

+        )

+        M = sp.Matrix([[1 / self.m_1, 0], [0, 1 / self.m_2]])

+        A = M * K

+

+        self.eigenvecs = []

+        self.eigenvals = []

+        for ev, mult, vecs in A.eigenvects():

+            self.eigenvecs.append(sp.Matrix(vecs))

+            self.eigenvals.extend([ev] * mult)

+

+    def __call__(self, t, x_0, k_1, k_c, k_2, m_1, m_2):

+        params = {

+            self.k_1: k_1,

+            self.k_c: k_c,

+            self.k_2: k_2,

+            self.m_1: m_1,

+            self.m_2: m_2,

+        }

+        x_0 = sp.Matrix(x_0)

+        eig_mat = sp.Matrix.hstack(*self.eigenvecs).subs(params)

+        g = eig_mat.inv() * x_0

+        L = sp.Matrix(

+            [

+                g[0] * sp.cos(self.eigenvals[0].subs(params) * self.t),

+                g[1] * sp.cos(self.eigenvals[1].subs(params) * self.t),

+            ]

+        )

+        x = eig_mat * L

+        f = sp.lambdify(self.t, x, "numpy")

+        return f(t).squeeze()

diff --git a/buch/papers/kra/teil0.tex b/buch/papers/kra/teil0.tex
deleted file mode 100644
index d06a055..0000000
--- a/buch/papers/kra/teil0.tex
+++ /dev/null
@@ -1,22 +0,0 @@
-%
-% einleitung.tex -- Beispiel-File für die Einleitung
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 0\label{kra:section:teil0}}
-\rhead{Teil 0}
-Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam
-nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam
-erat, sed diam voluptua \cite{kra:bibtex}.
-At vero eos et accusam et justo duo dolores et ea rebum.
-Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum
-dolor sit amet.
-
-Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam
-nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam
-erat, sed diam voluptua.
-At vero eos et accusam et justo duo dolores et ea rebum.  Stet clita
-kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit
-amet.
-
-
diff --git a/buch/papers/kra/teil1.tex b/buch/papers/kra/teil1.tex
deleted file mode 100644
index 0c0977d..0000000
--- a/buch/papers/kra/teil1.tex
+++ /dev/null
@@ -1,55 +0,0 @@
-%
-% teil1.tex -- Beispiel-File für das Paper
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 1
-\label{kra:section:teil1}}
-\rhead{Problemstellung}
-Sed ut perspiciatis unde omnis iste natus error sit voluptatem
-accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
-quae ab illo inventore veritatis et quasi architecto beatae vitae
-dicta sunt explicabo.
-Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit
-aut fugit, sed quia consequuntur magni dolores eos qui ratione
-voluptatem sequi nesciunt
-\begin{equation}
-\int_a^b x^2\, dx
-=
-\left[ \frac13 x^3 \right]_a^b
-=
-\frac{b^3-a^3}3.
-\label{kra:equation1}
-\end{equation}
-Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet,
-consectetur, adipisci velit, sed quia non numquam eius modi tempora
-incidunt ut labore et dolore magnam aliquam quaerat voluptatem.
-
-Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis
-suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur?
-Quis autem vel eum iure reprehenderit qui in ea voluptate velit
-esse quam nihil molestiae consequatur, vel illum qui dolorem eum
-fugiat quo voluptas nulla pariatur?
-
-\subsection{De finibus bonorum et malorum
-\label{kra:subsection:finibus}}
-At vero eos et accusamus et iusto odio dignissimos ducimus qui
-blanditiis praesentium voluptatum deleniti atque corrupti quos
-dolores et quas molestias excepturi sint occaecati cupiditate non
-provident, similique sunt in culpa qui officia deserunt mollitia
-animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}.
-
-Et harum quidem rerum facilis est et expedita distinctio
-\ref{kra:section:loesung}.
-Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil
-impedit quo minus id quod maxime placeat facere possimus, omnis
-voluptas assumenda est, omnis dolor repellendus
-\ref{kra:section:folgerung}.
-Temporibus autem quibusdam et aut officiis debitis aut rerum
-necessitatibus saepe eveniet ut et voluptates repudiandae sint et
-molestiae non recusandae.
-Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis
-voluptatibus maiores alias consequatur aut perferendis doloribus
-asperiores repellat.
-
-
diff --git a/buch/papers/kra/teil2.tex b/buch/papers/kra/teil2.tex
deleted file mode 100644
index 249f078..0000000
--- a/buch/papers/kra/teil2.tex
+++ /dev/null
@@ -1,40 +0,0 @@
-%
-% teil2.tex -- Beispiel-File für teil2 
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 2 
-\label{kra:section:teil2}}
-\rhead{Teil 2}
-Sed ut perspiciatis unde omnis iste natus error sit voluptatem
-accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
-quae ab illo inventore veritatis et quasi architecto beatae vitae
-dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit
-aspernatur aut odit aut fugit, sed quia consequuntur magni dolores
-eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam
-est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci
-velit, sed quia non numquam eius modi tempora incidunt ut labore
-et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima
-veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam,
-nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure
-reprehenderit qui in ea voluptate velit esse quam nihil molestiae
-consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla
-pariatur?
-
-\subsection{De finibus bonorum et malorum
-\label{kra:subsection:bonorum}}
-At vero eos et accusamus et iusto odio dignissimos ducimus qui
-blanditiis praesentium voluptatum deleniti atque corrupti quos
-dolores et quas molestias excepturi sint occaecati cupiditate non
-provident, similique sunt in culpa qui officia deserunt mollitia
-animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis
-est et expedita distinctio. Nam libero tempore, cum soluta nobis
-est eligendi optio cumque nihil impedit quo minus id quod maxime
-placeat facere possimus, omnis voluptas assumenda est, omnis dolor
-repellendus. Temporibus autem quibusdam et aut officiis debitis aut
-rerum necessitatibus saepe eveniet ut et voluptates repudiandae
-sint et molestiae non recusandae. Itaque earum rerum hic tenetur a
-sapiente delectus, ut aut reiciendis voluptatibus maiores alias
-consequatur aut perferendis doloribus asperiores repellat.
-
-
diff --git a/buch/papers/kra/teil3.tex b/buch/papers/kra/teil3.tex
deleted file mode 100644
index 2515c7d..0000000
--- a/buch/papers/kra/teil3.tex
+++ /dev/null
@@ -1,40 +0,0 @@
-%
-% teil3.tex -- Beispiel-File für Teil 3
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 3
-\label{kra:section:teil3}}
-\rhead{Teil 3}
-Sed ut perspiciatis unde omnis iste natus error sit voluptatem
-accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
-quae ab illo inventore veritatis et quasi architecto beatae vitae
-dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit
-aspernatur aut odit aut fugit, sed quia consequuntur magni dolores
-eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam
-est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci
-velit, sed quia non numquam eius modi tempora incidunt ut labore
-et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima
-veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam,
-nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure
-reprehenderit qui in ea voluptate velit esse quam nihil molestiae
-consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla
-pariatur?
-
-\subsection{De finibus bonorum et malorum
-\label{kra:subsection:malorum}}
-At vero eos et accusamus et iusto odio dignissimos ducimus qui
-blanditiis praesentium voluptatum deleniti atque corrupti quos
-dolores et quas molestias excepturi sint occaecati cupiditate non
-provident, similique sunt in culpa qui officia deserunt mollitia
-animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis
-est et expedita distinctio. Nam libero tempore, cum soluta nobis
-est eligendi optio cumque nihil impedit quo minus id quod maxime
-placeat facere possimus, omnis voluptas assumenda est, omnis dolor
-repellendus. Temporibus autem quibusdam et aut officiis debitis aut
-rerum necessitatibus saepe eveniet ut et voluptates repudiandae
-sint et molestiae non recusandae. Itaque earum rerum hic tenetur a
-sapiente delectus, ut aut reiciendis voluptatibus maiores alias
-consequatur aut perferendis doloribus asperiores repellat.
-
-
diff --git a/buch/papers/kreismembran/Makefile b/buch/papers/kreismembran/Makefile
index ce3c89f..a13f2cf 100644
--- a/buch/papers/kreismembran/Makefile
+++ b/buch/papers/kreismembran/Makefile
@@ -4,6 +4,6 @@
 # (c) 2020 Prof Dr Andreas Mueller
 #
 
-images:
-	@echo "no images to be created in kreismembran"
+images/TikzSaite.pdf: images/TikzSaite.tex
+	cd images && pdflatex TikzSaite.tex
 
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/TikzSaite.pdf b/buch/papers/kreismembran/images/TikzSaite.pdf
new file mode 100644
index 0000000..f95ceb9
--- /dev/null
+++ b/buch/papers/kreismembran/images/TikzSaite.pdf
diff --git a/buch/papers/kreismembran/images/TikzSaite.tex b/buch/papers/kreismembran/images/TikzSaite.tex
new file mode 100644
index 0000000..bf3d8f6
--- /dev/null
+++ b/buch/papers/kreismembran/images/TikzSaite.tex
@@ -0,0 +1,57 @@
+% vim: ts=2 sw=2 et :
+\documentclass[tikz, border=2mm]{standalone}
+
+\usepackage{times}
+\usepackage{txfonts}
+
+\begin{document}
+  \begin{tikzpicture}[
+      axis/.style = {very thick, -latex},
+      axis tick/.style = {
+        draw, draw = black, fill = black, rectangle,
+        inner sep = 0pt,
+        minimum height = 2mm,
+        minimum width = 1pt,
+      },
+      string/.style = {
+        ultra thick, draw = black,
+      },
+      string end/.style = {
+        string, circle, fill = gray,
+        inner sep = 0pt, minimum size = 1mm,
+      },
+      force/.style = {
+        very thick, draw = gray, -latex,
+      },
+    ]
+
+    % axes
+    \draw[axis] (0, 0) -- (8cm, 0) node[right] {$x$};
+    \draw[axis] (0, 0) -- (0, 5cm) node[above] {$u(x, t)$};
+
+    % axes ticks
+    \node[axis tick, label = {-90:$x_0$}] at (2cm, 0) {};
+    \node[axis tick, label = {-90:$x_0 + dx$}] at (6cm, 0) {};
+
+    % string
+    \coordinate (A) at (2cm, 2cm);
+    \coordinate (B) at (6cm, 4cm);
+
+    \draw[string] (A) to[out = 40, in = 200] (B);
+
+    \draw[force] (A) -- ++(220:15mm) node[gray, below right] {$T_1$};
+    \draw[force] (B) -- ++(20:15mm) node[gray, above left] {$T_2$};
+
+    \draw[dashed, gray, thick] (A) -- ++(-15mm, 0);
+    \draw[gray, thick] (A) ++ (-7mm,0) arc (180:220:7mm)
+      node[midway, left] {$\alpha$};
+
+    \draw[dashed, gray, thick] (B) -- ++(15mm, 0);
+    \draw[gray, thick] (B) ++ (7mm,0) arc (0:20:7mm)
+      node[pos = 0, below] {$\beta$};
+
+    \node[string end, label={110:$P_1$}] at (A) {};
+    \node[string end, label={110:$P_2$}] at (B) {};
+
+  \end{tikzpicture}
+\end{document}
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
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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
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+++ 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
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+++ 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
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+++ 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..65173f8 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},
@@ -81,4 +89,10 @@
 	type = {Dissertation},
 	author = {{Eric John Ruggiero Doctor of Philosophy In Mechanical Engineering}},
 	date = {2005},
+}
+
+@online{noauthor_laplace_nodate,
+	title = {Laplace Transform of Bessel Function of the First Kind of Order Zero - {ProofWiki}},
+	url = {https://proofwiki.org/wiki/Laplace_Transform_of_Bessel_Function_of_the_First_Kind_of_Order_Zero},
+	urldate = {2022-08-15},
 }
\ No newline at end of file
diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex
index bb8188d..e962aab 100644
--- a/buch/papers/kreismembran/teil0.tex
+++ b/buch/papers/kreismembran/teil0.tex
@@ -5,51 +5,59 @@
 %
 \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. 
-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 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 \dots''. 
+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 Membran 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. 
 
 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 mit einer Kraft $ T $ gespannt werden.
+	\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.
+	Die Membran wird also nicht durch ihr umliegendes Medium abgebremst noch erfährt sie Reibungsverluste durch Deformation.
 	
 \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.
+\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 dasselbe 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}
+		\includegraphics[]{papers/kreismembran/images/TikzSaite.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} $.
-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 
+In Abbildung \ref{kreismembran:im:Saite} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert.
+Wie für die Membran ist die Annahme iii) gültig, es entsteht keine Bewegung entlang der $ x $-Achse.
+Um dies zu erfüllen, muss der Punkt $ P_1 $ gleich stark entgegen der $ x $-Achse gezogen werden wie der Punkt $ P_2 $ in Richtung der $ x $-Achse 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*} 
@@ -57,7 +65,7 @@ befolgen. Die senkrecht wirkenden Kräfte werden mit $ T_1 $ und $ T_2 $ ausgedr
 \begin{equation*}
 	T_2 \sin \beta - T_1 \sin \alpha = \rho dx \frac{\partial^2 u}{\partial t^2} .
 \end{equation*}
-Die Gleichung wird durch $ T $ dividiert, wobei $ T $ nach \ref{kreismembran:eq:no_translation} geschickt gewählt wird. Somit kann
+Die Gleichung wird durch $ T $ dividiert, wobei $ T $ nach \eqref{kreismembran:eq:no_translation} geschickt gewählt wird. Somit kann
 \begin{equation*}
 	\frac{T_2 \sin \beta}{T_2 \cos \beta} - \frac{T_1 \sin \alpha}{T_1 \cos \alpha} = \frac{\rho dx}{T} \frac{\partial^2 u}{\partial t^2}
 \end{equation*}
@@ -69,14 +77,19 @@ 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.
+Wenn $ dx $ als unendlich kleines Stück betrachtet wird, ergibt sich als Grenzwert die zweite Ableitung von $ u(x,t) $ nach $ x $. 
+Der Term $ \frac{\rho}{T} $ wird durch $ c^2 $ ersetzt, da der Bruch für eine gegebene Membran eine positive Konstante sein muss. 
+Damit 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 verwendet werden.
\ No newline at end of file
diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex
index 39ca598..a9b2fad 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:
@@ -23,90 +23,82 @@ Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so d
 	\frac1r
 	\frac{\partial}{\partial r}
 	+
-	\frac{1}{r 2}
+	\frac{1}{r^2}
 	\frac{\partial^2}{\partial\varphi^2}
 	\label{buch:pde:kreis:laplace}
 \end{equation*}
 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 Abschnitt \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 Separationsmethode gelöst.
 \subsubsection{Ansatz der Separation der Variablen\label{subsub:ansatz_separation}}
-Daher muss an dieser Stelle von einer Separation der Variablen ausgegangen werden:
+Hierfür wird folgenden Ansatz gemacht:
 \begin{equation*}
-	u(r,\varphi, t) = F(r)G(\varphi)T(t)
+	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 nach Division durch $u$:
 \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)}.
 \end{align*}
-In der zweiten Gleichung hängt die linke Seite nur von $r$ ab, während die rechte Seite nur von $\varphi$ abhängt. Sie müssen also wiederum gleich einer reellen Zahl $\nu$ sein. Also das:
+In der zweiten Gleichung hängt die linke Seite nur von $r$ ab, während die rechte Seite nur von $\varphi$ abhängt. Sie müssen also wiederum gleich einer reellen Zahl $\nu$ sein. Also:
 \begin{align*}
-	r^2F''(r) + rF'(r) + (\kappa^2 r^2 - \nu)F(r) &= 0 \\
-	G''(\varphi) &= \nu G(\varphi).
+	r^2F''(r) + rF'(r) + (\kappa^2 r^2 - \nu)F(r) = 0 \quad \text{und} \quad
+	G''(\varphi) = \nu G(\varphi).
 \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 $G''(\varphi)=-n^2 G(\varphi)$ gilt, schreibt man die gemeinsame Konstante als $\nu=-n^2$, was die Formeln später vereinfacht. $n$ muss auch eine ganze Zahl sein, weil  $G(\varphi)$ sonst nicht $2\pi$-periodisch ist. Also:
 \begin{equation*}
- G(\varphi) = C_n \cos(\varphi) + D_n \sin(\varphi)
+ G(\varphi) = C_n \cos(n\varphi) + D_n \sin(n\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 \label{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}
 \end{align}
-Wir bereits in Kapitel \ref{buch:differntialgleichungen:section:bessel} gezeigt, sind die Besselfunktionen
+Wie bereits in Kapitel \ref{buch:differntialgleichungen:section:bessel} gezeigt, sind die Bessel-Funktionen
 \begin{equation*}
-	J_{\nu}(x) = r^\nu \displaystyle\sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m+\nu}m! \Gamma (\nu + m+1)}
+	J_{n}(x) = r^n \displaystyle\sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m+n}m! \Gamma (n + m+1)}
 \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 - n^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 $G(\varphi)$. 
-
+Die Differenzialgleichung $T''(t) + c^2\kappa^2T(t) = 0$, wird auf ähnliche Weise gelöst wie $G(\varphi)$. Um eine Einschränkung der möglichen Frequenzen zu erhalten und die Lösung als Reihe schreiben zu können, muss die folgende homogene Randbedingung definiert werden: 
+\begin{equation*}
+	u\big|_{\Gamma} = 0 \quad \text{für} \quad 0 \leq \varphi \leq 2\pi,\quad t \geq 0,
+\end{equation*}
+welche die $\kappa$ auf mögliche werte $\kappa_{mn}$ einschränkt.
 \subsubsection{Zusammenfassung der Lösungen\label{subsub:zusammenfassung_lösungen}}
 Durch Überlagerung aller Ergebnisse erhält man die Lösung
 \begin{align}
@@ -115,7 +107,23 @@ 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}). $J_n(\kappa_{mn}r)$ 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.
+\begin{center}
+	* \quad *\quad *
+\end{center}
+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..4ceeb84 100644
--- a/buch/papers/kreismembran/teil2.tex
+++ b/buch/papers/kreismembran/teil2.tex
@@ -7,34 +7,34 @@
 
 Hermann Hankel (1839--1873) war ein deutscher Mathematiker, der für seinen Beitrag zur mathematischen Analysis und insbesondere für die nach ihm benannte Transformation bekannt ist.
 Diese Transformation tritt bei der Untersuchung von Funktionen auf, die nur von der Entfernung des Ursprungs abhängen.
-Er studierte auch Funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der dritten Art.
+Er untersuchte auch Funktionen, jetzt Hankel- oder Bessel-Funktionen genannt, der dritten Art.
 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}}
-Wir führen die Definition der Hankel-Transformation \cite{lokenath_debnath_integral_2015} aus der zweidimensionalen Fourier-Transformation und ihrer Umkehrung ein, die durch:
+\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-Trans\-formation 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:
+definiert ist, wobei $\mathbf{r}=(x,y)$ 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 Bessel-Funktion 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),
@@ -47,49 +47,40 @@ 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:
+Die inverse \textit{Hankel-Transformation} ist also als 
 \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}
+definiert.
 
-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, 
+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}}
-In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation aufgeführt. Der Beweis für ihre Gültigkeit wird jedoch nicht analysiert.
+\subsection{Operatoreigenschaften der Hankel-Transformation \label{sub:op_properties_hankel}}
+In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation aufgeführt. Die Beweise für ihre Gültigkeit werden jedoch nicht analysiert, diese sind im Buch \textit{Integral Tansforms and Their Applications} \cite{lokenath_debnath_integral_2015} zu finden.
 
 \begin{satz}{Skalierung:}
-	Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann:
+	Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann gilt:
 	
 	\begin{equation*}
 		\mathscr{H}_n\{f(ar)\}=\frac{1}{a^{2}}\tilde{f}_n \left(\frac{\kappa}{a}\right), \quad a>0.
 	\end{equation*}
 \end{satz}
 
-\begin{satz}{Persevalsche Relation (Skalarprodukt bleibt erhalten):}
-Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann:
+\begin{satz}{Parsevalsche Relation:}
+Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann gilt:
 
 \begin{equation*}
 	\int_{0}^{\infty}rf(r)g(r) \; dr = \int_{0}^{\infty}\kappa\tilde{f}(\kappa)\tilde{g}(\kappa) \; d\kappa.
@@ -97,20 +88,20 @@ Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H
 \end{satz}
 
 \begin{satz}{Hankel-Transformationen von Ableitungen:}
-Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann:
+Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann gilt:
 
 \begin{align*}
 	&\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}
-Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann:
+Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann gilt:
 
 \begin{equation*}
 	\mathscr{H}_n \left\{ \left( \nabla^2 - \frac{n^2}{r^2} f(r)\right)\right\}= \mathscr{H}_n\left\{\frac{1}{r}\frac{d}{dr}\left(r\frac{df}{dr}\right) - \frac{n^2}{r^2}f(r)\right\}=-\kappa^2\tilde{f}_{n}(\kappa),
 \end{equation*}
-bereitgestellt dass $rf'(r)$ und $rf(r)$ verschwinden für $r\to0$ und $r\to\infty$.
+bereitgestellt, dass $rf'(r)$ und $rf(r)$ verschwinden für $r\to0$ und $r\to\infty$.
 \end{satz}
diff --git a/buch/papers/kreismembran/teil3.tex b/buch/papers/kreismembran/teil3.tex
index 7d5648a..d143ec7 100644
--- a/buch/papers/kreismembran/teil3.tex
+++ b/buch/papers/kreismembran/teil3.tex
@@ -6,25 +6,22 @@
 \section{Lösungsmethode 2: Transformationsmethode
 \label{kreismembran:section:teil3}}
 \rhead{Lösungsmethode 2: Transformationsmethode}
-Die Hankel-Transformation wird dann zur Lösung der Differentialgleichung verwendet. Es müssen jedoch einige Änderungen an dem Problem vorgenommen werden, damit es mit den Annahmen übereinstimmt, die für die Verwendung der Hankel-Transformation erforderlich sind. Das heisst, dass die Funktion $u$ nur von der Entfernung zum Ausgangspunkt abhängt. 
+Die Hankel-Transformation kann hier zur Lösung der Differentialgleichung verwendet werden. Es müssen jedoch einige Änderungen an dem Problem vorgenommen werden, damit es mit den Annahmen übereinstimmt, die für die Verwendung der Hankel-Transformation erforderlich sind. Das heisst, dass die Funktion $u$ nur von der Entfernung zum Ausgangspunkt abhängt. 
 
 \subsubsection{Transformation und Reduktion auf eine algebraische Gleichung\label{subsub:transf_reduktion}}
 Führt man also das Konzept einer unendlichen und achsensymmetrischen Membran ein:
-\begin{equation*}
+\begin{align}
 	\frac{\partial^2u}{\partial t^2}
 	=
 	c^2  \left(\frac{\partial^2 u}{\partial r^2}
 	+
 	\frac{1}{r}
-	\frac{\partial u}{\partial r} \right), \quad 0<r<\infty, \quad t>0
-	\label{eq:PDE_inf_membane}
-\end{equation*}
-
-\begin{align}
-	u(r,0)=f(r), \quad u_t(r,0) = g(r), \quad \text{für} \quad 0<r<\infty
+	\frac{\partial u}{\partial r} \right), \quad 0<r<\infty, \quad t>0   \label{eq:PDE_inf_membane} \\
+		u(r,0)=f(r), \quad u_t(r,0) = g(r), \quad \text{für} \quad 0<r<\infty.
 	\label{eq:PDE_inf_membane_RB}
 \end{align}
 
+
 Mit Anwendung der Hankel-Transformation nullter Ordnung in Abhängigkeit von $r$ auf die Gleichungen \eqref{eq:PDE_inf_membane} und \eqref{eq:PDE_inf_membane_RB}:
 
 \begin{align}
@@ -40,12 +37,12 @@ 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).
 \end{equation*}
-Wendet man an nun die inverse Hankel-Transformation an, so erhält man die formale Lösung
+Wendet man nun die inverse Hankel-Transformation an, so erhält man die formale Lösung
 
 \begin{align}
 	u(r,t)=\int_{0}^{\infty}\kappa\tilde{f}(\kappa)\cos(c\kappa t) J_0(\kappa r) \; d\kappa +\frac{1}{c}\int_{0}^{\infty}\tilde{g}(\kappa)\sin(c\kappa t)J_0(\kappa r) \; d\kappa.
@@ -53,22 +50,33 @@ Wendet man an nun die inverse Hankel-Transformation an, so erhält man die forma
 \end{align}
 
 \subsubsection{Erfüllung der Anfangsbedingungen\label{subsub:erfüllung_AB}}
-Es wird in Folgenden davon ausgegangen, dass sich die Membran verformt und zum Zeitpunkt $t=0$ freigegeben wird
+Es wird im Folgenden davon ausgegangen, dass sich die Membran verformt und zum Zeitpunkt $t=0$ freigegeben wird
 
 \begin{equation*}
 	u(r,0)=f(r)=Aa(r^2 + a^2)^{-\frac{1}{2}}, \quad u_t(r,0)=g(r)=0
 \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
+
+\noindent Die formale Lösung  \eqref{eq:formale_lösung} lautet also
+\begin{align}
+	u(r,t)=Aa\int_{0}^{\infty}e^{-a\kappa} J_0(\kappa r)\cos(c\kappa t) \; dk=AaRe\int_{0}^{\infty}e^{-\kappa(a+ict)} J_0(\kappa r) \; dk.
+	\label{form_lösung2_step1}
+\end{align}
+
+\noindent Aus der Laplace-Transformation und unter Verwendung der Skalierungseigenschaft \cite{noauthor_laplace_nodate} ergibt sich, dass
 \begin{align*}
-	u(r,t)&=Aa\int_{0}^{\infty}e^{-a\kappa} J_0(\kappa r)\cos(c\kappa t) \; dk=AaRe\int_{0}^{\infty}e^{-\kappa(a+ict)} J_0(\kappa r) \; dk\\
-	&=AaRe\left\{r^2+\left(a+ict\right)^2\right\}^{-\frac{1}{2}}
-\end{align*}
+	\int_{0}^{\infty}e^{-px} J_0(\kappa x) \; dx = \frac{1}{\sqrt{\kappa^2 + p^2}},
+\end{align*}	
+
+\noindent \eqref{form_lösung2_step1} kann somit vereinfacht werden in:
+\begin{equation*}
+	u(r,t)=AaRe\left\{r^2+\left(a+ict\right)^2\right\}^{-\frac{1}{2}}.
+\end{equation*}
 
-Nimmt man jedoch die allgemeine Lösung mit Summationen, 
+\noindent 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 +86,8 @@ 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.
-Die Funktion hängt also nicht mehr von der Besselfunktionen $n$-ter Ordnung ab, sondern nur von der $0$-ter Ordnung. 
+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 Bessel-Funktionen $n$-ter Ordnung ab, sondern nur von der nullter Ordnung. 
 
 
diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex
index c124354..d6aa54f 100644
--- a/buch/papers/kreismembran/teil4.tex
+++ b/buch/papers/kreismembran/teil4.tex
@@ -5,12 +5,190 @@
 %
 \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} $ steht 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 $ ist 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 $ die Anzahl von Zeitschritten ist.
+Das Element auf Zeile $ i $, Spalte $ j $ der $ w $-ten Matrix von $ U[] $ also $ U[w]_{ij} $ ist somit die 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-Elemente 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 plus 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. Dieses Verfahren wird Euler-Methode genannt.
+
+\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.
+\subsubsection{Initialisierung}
+Die Anzahl der simulierten Elemente soll $ m \times n $ sein, was 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 einem Tiefpassfilter in der Bildverarbeitung entspricht.
+
+\subsubsection{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, dass 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 $0$ sein.
+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])\odot M\\
+	\label{kreismembran:eq:folge_V}
+	V[w+1] &= (V[w] + dt \cdot \Delta_h u \cdot c^2)\odot M
+\end{align}
+berechnet werden. Das Symbol $\odot$ steht hier für eine elementweise Matrixmultiplikation (Hadamard-Produkt)
+\subsubsection{Simulation}
+Mit den gegebenen Gleichungen \eqref{kreismembran:eq:folge_U} und \eqref{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 Zentrum beeinflusst.
+Soll erst gar keine Reflexion entstehen, muss ein Absorber modelliert werden welcher die Störung möglichst ohne Reflexion aufnimmt.
+
+\subsubsection{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 $M$ einen binären 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 ist 
+\begin{equation*}
+	r(i,j) = \sqrt{|i-\frac{m}{2}|^2+|j-\frac{n}{2}|^2},
+\end{equation*} 
+wobei $ m $ und $n$ die Dimensionen der Matrix sind.
+Für einen stufenlosen Übergang werden die Elemente der Maske auf 
+
+\begin{align}
+	M_{ij} = \begin{cases} 1-e^{(r(i,j)-b)a} & \text{$x > b$} \\
+		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} 
+\subsubsection{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 weiss, 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/kugel/Makefile b/buch/papers/kugel/Makefile
index f798a55..995206b 100644
--- a/buch/papers/kugel/Makefile
+++ b/buch/papers/kugel/Makefile
@@ -5,5 +5,6 @@
 #
 
 images:
-	@echo "no images to be created in kugel"
+	$(MAKE) -C ./figures/povray/
+	$(MAKE) -C ./figures/tikz/
 
diff --git a/buch/papers/kugel/figures/flux.pdf b/buch/papers/kugel/figures/flux.pdf
new file mode 100644
index 0000000..6a87288
--- /dev/null
+++ b/buch/papers/kugel/figures/flux.pdf
diff --git a/buch/papers/kugel/images/Makefile b/buch/papers/kugel/figures/povray/Makefile
index 4226dab..4226dab 100644
--- a/buch/papers/kugel/images/Makefile
+++ b/buch/papers/kugel/figures/povray/Makefile
diff --git a/buch/papers/kugel/figures/povray/curvature.jpg b/buch/papers/kugel/figures/povray/curvature.jpg
new file mode 100644
index 0000000..6448966
--- /dev/null
+++ b/buch/papers/kugel/figures/povray/curvature.jpg
diff --git a/buch/papers/kugel/images/curvature.maxima b/buch/papers/kugel/figures/povray/curvature.maxima
index 6313642..6313642 100644
--- a/buch/papers/kugel/images/curvature.maxima
+++ b/buch/papers/kugel/figures/povray/curvature.maxima
diff --git a/buch/papers/kugel/figures/povray/curvature.png b/buch/papers/kugel/figures/povray/curvature.png
new file mode 100644
index 0000000..20268f2
--- /dev/null
+++ b/buch/papers/kugel/figures/povray/curvature.png
diff --git a/buch/papers/kugel/images/curvature.pov b/buch/papers/kugel/figures/povray/curvature.pov
index 3b15d77..3b15d77 100644
--- a/buch/papers/kugel/images/curvature.pov
+++ b/buch/papers/kugel/figures/povray/curvature.pov
diff --git a/buch/papers/kugel/images/curvgraph.m b/buch/papers/kugel/figures/povray/curvgraph.m
index 75effd6..75effd6 100644
--- a/buch/papers/kugel/images/curvgraph.m
+++ b/buch/papers/kugel/figures/povray/curvgraph.m
diff --git a/buch/papers/kugel/images/spherecurve.cpp b/buch/papers/kugel/figures/povray/spherecurve.cpp
index 8ddf5e5..8ddf5e5 100644
--- a/buch/papers/kugel/images/spherecurve.cpp
+++ b/buch/papers/kugel/figures/povray/spherecurve.cpp
diff --git a/buch/papers/kugel/figures/povray/spherecurve.jpg b/buch/papers/kugel/figures/povray/spherecurve.jpg
new file mode 100644
index 0000000..cd2e7c8
--- /dev/null
+++ b/buch/papers/kugel/figures/povray/spherecurve.jpg
diff --git a/buch/papers/kugel/images/spherecurve.m b/buch/papers/kugel/figures/povray/spherecurve.m
index 99d5c9a..99d5c9a 100644
--- a/buch/papers/kugel/images/spherecurve.m
+++ b/buch/papers/kugel/figures/povray/spherecurve.m
diff --git a/buch/papers/kugel/images/spherecurve.maxima b/buch/papers/kugel/figures/povray/spherecurve.maxima
index 1e9077c..1e9077c 100644
--- a/buch/papers/kugel/images/spherecurve.maxima
+++ b/buch/papers/kugel/figures/povray/spherecurve.maxima
diff --git a/buch/papers/kugel/figures/povray/spherecurve.png b/buch/papers/kugel/figures/povray/spherecurve.png
new file mode 100644
index 0000000..ff24371
--- /dev/null
+++ b/buch/papers/kugel/figures/povray/spherecurve.png
diff --git a/buch/papers/kugel/images/spherecurve.pov b/buch/papers/kugel/figures/povray/spherecurve.pov
index b1bf4b8..b1bf4b8 100644
--- a/buch/papers/kugel/images/spherecurve.pov
+++ b/buch/papers/kugel/figures/povray/spherecurve.pov
diff --git a/buch/papers/kugel/figures/tikz/Makefile b/buch/papers/kugel/figures/tikz/Makefile
new file mode 100644
index 0000000..4ec4e5a
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/Makefile
@@ -0,0 +1,12 @@
+FIGURES := spherical-coordinates.pdf curvature-1d.pdf
+
+all: $(FIGURES)
+
+%.pdf: %.tex
+	pdflatex $<
+
+curvature-1d.pdf: curvature-1d.tex curvature-1d.dat
+	pdflatex curvature-1d.tex
+
+curvature-1d.dat: curvature-1d.py
+	python3 $<
diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.dat b/buch/papers/kugel/figures/tikz/curvature-1d.dat
new file mode 100644
index 0000000..6622398
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/curvature-1d.dat
@@ -0,0 +1,500 @@
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diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.pdf b/buch/papers/kugel/figures/tikz/curvature-1d.pdf
new file mode 100644
index 0000000..6425af6
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/curvature-1d.pdf
diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.py b/buch/papers/kugel/figures/tikz/curvature-1d.py
new file mode 100644
index 0000000..4710fc8
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/curvature-1d.py
@@ -0,0 +1,32 @@
+import numpy as np
+import matplotlib.pyplot as plt
+
+
+@np.vectorize
+def fn(x):
+    return (x ** 2) * 2 / 100 + (1 + x / 4) + np.sin(x)
+
+@np.vectorize
+def ddfn(x):
+    return 2 * 5 / 100 - np.sin(x)
+
+x = np.linspace(0, 10, 500)
+y = fn(x)
+ddy = ddfn(x)
+
+cmap = ddy - np.min(ddy)
+cmap = cmap * 1000 / np.max(cmap)
+
+plt.plot(x, y)
+plt.plot(x, ddy)
+# plt.plot(x, cmap)
+
+plt.show()
+
+fname = "curvature-1d.dat"
+np.savetxt(fname, np.array([x, y, cmap]).T, delimiter="  ")
+
+# with open(fname, "w") as f:
+#     # f.write("x   y   cmap\n")
+#     for xv, yv, cv in zip(x, y, cmap):
+#         f.write(f"{xv}   {yv}   {cv}\n")
diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.tex b/buch/papers/kugel/figures/tikz/curvature-1d.tex
new file mode 100644
index 0000000..6983fb0
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/curvature-1d.tex
@@ -0,0 +1,21 @@
+% vim:ts=2 sw=2 et:
+\documentclass[tikz, border=5mm]{standalone}
+\usepackage{pgfplots}
+
+\begin{document}
+\begin{tikzpicture}
+  \begin{axis}[
+      clip = false,
+      width = 8cm, height = 6cm,
+      xtick = \empty, ytick = \empty,
+      colormap name = viridis,
+      axis lines = middle,
+      axis line style = {ultra thick, -latex}
+    ]
+    \addplot+[
+      smooth, mark=none, line width = 3pt, mesh,
+      point meta=explicit,
+    ] file {curvature-1d.dat};
+  \end{axis}
+\end{tikzpicture}
+\end{document}
diff --git a/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf b/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf
new file mode 100644
index 0000000..1bff016
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf
diff --git a/buch/papers/kugel/figures/tikz/spherical-coordinates.tex b/buch/papers/kugel/figures/tikz/spherical-coordinates.tex
new file mode 100644
index 0000000..3a45385
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/spherical-coordinates.tex
@@ -0,0 +1,99 @@
+\documentclass[tikz]{standalone}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{bm}
+\usepackage{lmodern}
+\usepackage{tikz-3dplot}
+
+\usetikzlibrary{arrows}
+\usetikzlibrary{intersections}
+\usetikzlibrary{math}
+\usetikzlibrary{positioning}
+\usetikzlibrary{arrows.meta}
+\usetikzlibrary{shapes.misc}
+\usetikzlibrary{calc}
+
+\begin{document}
+
+\tdplotsetmaincoords{60}{130}
+\pgfmathsetmacro{\l}{2}
+
+\begin{tikzpicture}[
+    >=latex,
+    tdplot_main_coords,
+    dot/.style = {
+      black, fill = black, circle,
+      outer sep = 0, inner sep = 0,
+      minimum size = .8mm 
+    },
+    round/.style = {
+      draw = orange, thick, circle,
+      minimum size = 1mm,
+      inner sep = 0pt, outer sep = 0pt,
+    },
+    cross/.style = {
+      cross out, draw = magenta, thick,
+      minimum size = 1mm, 
+      inner sep = 0pt, outer sep = 0pt
+    },
+  ]
+
+  % origin
+  \coordinate (O) at (0,0,0);
+
+  % poles
+  \coordinate (NP) at (0,0,\l);
+  \coordinate (SP) at (0,0,-\l);
+  
+  % \draw (SP) node[dot, gray] {};
+  % \draw (NP) node[dot, gray] {};
+
+  % gray unit circle
+  \tdplotdrawarc[gray]{(O)}{\l}{0}{360}{}{};
+  \draw[gray, dashed] (-\l, 0, 0) to (\l, 0, 0);
+  \draw[gray, dashed] (0, -\l, 0) to (0, \l, 0);
+
+  % axis
+  \draw[->] (O) -- ++(1.25*\l,0,0) node[left] {\(\mathbf{\hat{x}}\)};
+  \draw[->] (O) -- ++(0,1.25*\l,0) node[right] {\(\mathbf{\hat{y}}\)};
+  \draw[->] (O) -- ++(0,0,1.25*\l) node[above] {\(\mathbf{\hat{z}}\)};
+
+  % meridians
+  \foreach \phi in {0, 30, 60, ..., 150}{
+    \tdplotsetrotatedcoords{\phi}{90}{0};
+    \tdplotdrawarc[lightgray, densely dotted, tdplot_rotated_coords]{(O)}{\l}{0}{360}{}{};
+  }
+
+  % dot above and its projection
+  \pgfmathsetmacro{\phi}{120}
+  \pgfmathsetmacro{\theta}{40}
+
+  \pgfmathsetmacro{\px}{cos(\phi)*sin(\theta)*\l}
+  \pgfmathsetmacro{\py}{sin(\phi)*sin(\theta)*\l}
+  \pgfmathsetmacro{\pz}{cos(\theta)*\l})
+  
+  % point A
+  \coordinate (A) at (\px,\py,\pz);
+  \coordinate (Ap) at (\px,\py, 0);
+  
+  % lines
+  \draw[red!80!black, ->] (O) -- (A);
+	\draw[red!80!black, densely dashed] (O) -- (Ap) -- (A)
+		node[above right] {\(\mathbf{\hat{r}}\)};
+  
+  % arcs
+  \tdplotdrawarc[blue!80!black, ->]{(O)}{.8\l}{0}{\phi}{}{};
+  \node[below right, blue!80!black] at (.8\l,0,0) {\(\bm{\hat{\varphi}}\)};
+  
+  \tdplotsetrotatedcoords{\phi-90}{-90}{0};
+  \tdplotdrawarc[blue!80!black, ->, tdplot_rotated_coords]{(O)}{.95\l}{0}{\theta}{}{};
+  \node[above right = 1mm, blue!80!black] at (0,0,.8\l) {\(\bm{\hat{\vartheta}}\)};
+
+
+  % dots
+  \draw (O) node[dot] {};
+  \draw (A) node[dot, fill = red!80!black] {};
+
+\end{tikzpicture}
+\end{document}
+% vim:ts=2 sw=2 et:
diff --git a/buch/papers/kugel/main.tex b/buch/papers/kugel/main.tex
index 98d9cb2..d063f87 100644
--- a/buch/papers/kugel/main.tex
+++ b/buch/papers/kugel/main.tex
@@ -14,6 +14,7 @@
 \input{papers/kugel/preliminaries}
 \input{papers/kugel/spherical-harmonics}
 \input{papers/kugel/applications}
+\input{papers/kugel/proofs}
 
 \printbibliography[heading=subbibliography]
 \end{refsection}
diff --git a/buch/papers/kugel/packages.tex b/buch/papers/kugel/packages.tex
index 61f91ad..ead7653 100644
--- a/buch/papers/kugel/packages.tex
+++ b/buch/papers/kugel/packages.tex
@@ -1,3 +1,4 @@
+% vim:ts=2 sw=2 et:
 %
 % packages.tex -- packages required by the paper kugel
 %
@@ -7,4 +8,13 @@
 % if your paper needs special packages, add package commands as in the
 % following example
 %\usepackage{packagename}
+\usepackage{cases}
 
+\newcommand{\kugeltodo}[1]{\textcolor{red!70!black}{\texttt{[TODO: #1]}}}
+\newcommand{\kugelplaceholderfig}[2]{ \begin{tikzpicture}%
+    \fill[lightgray!20] (0, 0) rectangle (#1, #2);%
+    \node[gray, anchor = center] at ({#1 / 2}, {#2 / 2}) {\Huge \ttfamily \bfseries TODO};
+  \end{tikzpicture}}
+
+\DeclareMathOperator{\sphlaplacian}{\nabla^2_{\mathit{S}}}
+\DeclareMathOperator{\surflaplacian}{\nabla^2_{\partial \mathit{S}}}
diff --git a/buch/papers/kugel/preliminaries.tex b/buch/papers/kugel/preliminaries.tex
index 03cd421..e48abe4 100644
--- a/buch/papers/kugel/preliminaries.tex
+++ b/buch/papers/kugel/preliminaries.tex
@@ -44,23 +44,23 @@ numbers \(\mathbb{R}\).
   \)
 \end{definition}
 
-\texttt{TODO: Text here.}
+\kugeltodo{Text here.}
 
 \begin{definition}[Span]
 \end{definition}
 
-\texttt{TODO: Text here.}
+\kugeltodo{Text here.}
 
 \begin{definition}[Linear independence]
 \end{definition}
 
 
-\texttt{TODO: Text here.}
+\kugeltodo{Text here.}
 
 \begin{definition}[Basis]
 \end{definition}
 
-\texttt{TODO: Text here.}
+\kugeltodo{Text here.}
 
 \begin{definition}[Inner product]
   \label{kugel:def:inner-product} \nocite{axler_linear_2014}
diff --git a/buch/papers/kugel/proofs.tex b/buch/papers/kugel/proofs.tex
new file mode 100644
index 0000000..143caa8
--- /dev/null
+++ b/buch/papers/kugel/proofs.tex
@@ -0,0 +1,245 @@
+% vim:ts=2 sw=2 et spell tw=80:
+\section{Proofs}
+
+\subsection{Legendre Functions} \label{kugel:sec:proofs:legendre}
+
+\kugeltodo{Fix theorem numbers to match, review text.}
+
+\begin{lemma}
+    The polynomial function
+    \begin{align*}
+        y_n(x)&=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k}\\
+        &= \frac{1}{n!2^n}\frac{d^n}{dx^n}(1-x^2)^n =: P_n(x),
+    \end{align*}
+    is a solution to the second order differential equation
+    \begin{equation}\label{kugel:eq:sol_leg}
+        (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx} + n(n+1)y=0, \quad \forall n>0.
+    \end{equation}
+\end{lemma}
+\begin{proof}
+    In order to find a solution to Eq.\eqref{eq:legendre}, the following Ansatz can be performed:
+    \begin{equation}\label{eq:ansatz}
+    y(x) = \sum_{k=0}^\infty a_k x^k.
+    \end{equation}
+    Given Eq.\eqref{eq:ansatz}, then
+    \begin{align*}
+    \frac{dy}{dx} &= \sum_{k=0}^\infty k a_k x^{k-1}, \\
+    \frac{d^2y}{dx^2} &= \sum_{k=0}^\infty k (k-1) a_k x^{k-2}.
+    \end{align*}
+    Eq.\eqref{eq:legendre} can be therefore written as
+    \begin{align}
+    &(1-x^2)\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0 \label{eq:ansatz_in_legendre} \\
+    &=\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0. \nonumber
+    \end{align}
+    If one consider the term
+    \begin{equation}\label{eq:term}
+    \sum_{k=0}^\infty k (k-1) a_k x^{k-2},
+    \end{equation}
+    the substitution $\tilde{k}=k-2$ yields Eq.\eqref{eq:term} to
+    \begin{equation*}
+    \sum_{\tilde{k}=-2}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}+2} x^{\tilde{k}}=\sum_{\tilde{k}=0}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}} x^{\tilde{k}}.
+    \end{equation*}
+    This means that Eq.\eqref{eq:ansatz_in_legendre} becomes
+    \begin{align}
+    &\sum_{k=0}^\infty (k+1)(k+2) a_{k+2} x^{k} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2\sum_{k=0}^\infty k a_k x^k + n(n+1)\sum_{k=0}^\infty a_k x^k \nonumber \\
+    = &\sum_{k=0}^\infty \big[ (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k \big] x^k \stackrel{!}{=} 0. \label{eq:condition}
+    \end{align}
+    The condition in Eq.\eqref{eq:condition} is equivalent to 
+    \begin{equation}\label{eq:condition_2}
+    (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k = 0.
+    \end{equation}
+    We can derive a recursion formula for $a_{k+2}$ from Eq.\eqref{eq:condition_2}, which can be expressed as
+    \begin{equation}\label{eq:recursion}
+    a_{k+2}= \frac{k (k-1) - 2 k + n(n+1)}{(k+1)(k+2)}a_k = \frac{(k-n)(k+n+1)}{(k+2)(k+1)}a_k.
+    \end{equation}
+    All coefficients can be calculated using the latter. 
+    
+    Following Eq.\eqref{eq:recursion}, if we want to compute $a_6$ we would have
+    \begin{align*}
+    a_{6}= -\frac{(n-4)(n+5)}{6\cdot 5}a_4 &= -\frac{(n-4)(5+n)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3}  a_2 \\
+    &= -\frac{(n-4)(n+5)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3} -\frac{n(n+1)}{2 \cdot 1}  a_0 \\
+    &= -\frac{(n+5)(n+3)(n+1)n(n-2)(n-4)}{6!} a_0.
+    \end{align*}
+    One can generalize this relation for the $i^\text{th}$ even coefficient as
+    \begin{equation*}
+    a_{2k} = (-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots (n-(2k-2)+2)(n-(2k-2))}{(2k)!}a_0
+    \end{equation*}
+    where $i=2k$.
+    
+    A similar expression can be written for the odd coefficients $a_{2k-1}$. In this case, the equation starts from $a_1$ and to find the pattern we can write the recursion for an odd coefficient, $a_7$ for example
+    \begin{align*}
+    a_{7}= -\frac{(n-5)(n+6)}{7\cdot 6}a_5 &= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4}  a_3 \\
+    &= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4}  -\frac{(n-1)(n+2)}{3 \cdot 2}  a_1 \\
+    &= -\frac{(n+6)(n+4)(n+2)(n-1)(n-3)(n-5)}{7!} a_1.
+    \end{align*}
+    As before, we can generalize this equation for the $i^\text{th}$ odd coefficient
+    \begin{equation*}
+    a_{2k+1} = (-1)^k \frac{(n + 2k)(n+2k-2)\hdots(n-(2k-1)+2)(n-(2k-1))}{(2k+1)!}a_1
+    \end{equation*}
+    where $i=2k+1$.
+    
+    Let be 
+    \begin{align*}
+    y_\text{e}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots \color{red}(n-(2k-2)+2)(n-(2k-2))}{(2k)!} x^{2k}, \\
+    y_\text{o}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n + 2k)(n+2k-2)\hdots \color{blue} (n-(2k-1)+2)(n-(2k-1))}{(2k+1)!} x^{2k+1}.
+    \end{align*}
+    The solution to the Eq.\eqref{eq:legendre} can be written as
+    \begin{equation}\label{eq:solution}
+    y(x) = \lim_{K \to \infty} \left[ a_0 y_\text{e}^K(x) + a_1 y_\text{o}^K(x) \right].
+    \end{equation}
+    
+    The colored parts can be analyzed separately:
+    \begin{itemize}
+      \item[\textcolor{red}{\textbullet}] Suppose that $n=n_0$ is an even number. Then the red part, for a specific value of $k=k_0$, will follow the following relation:
+    \begin{equation*}
+    n_0-(2k_0-2)=0. 
+    \end{equation*}
+    From that point on, given the recursive nature of Eq.\eqref{eq:recursion}, all the subsequent coefficients will also be 0, making the sum finite.
+    \begin{equation*}
+    a_{2k}=0 \iff y_{\text{o}}^{2k}(x)=y_{\text{o}}^{2k_0}(x), \quad \forall k>k_0
+    \end{equation*} 
+      \item[\textcolor{blue}{\textbullet}] Suppose that $n=n_0$ is an odd number. Then the blue part, for a specific value of $k=k_0$, will follow the following relation 
+    \begin{equation*}
+    n_0-(2k_0-1)=0.  
+    \end{equation*}
+    From that point on, for the same reason as before, all the subsequent coefficients will also be 0, making the sum finite.
+    \begin{equation*}
+    a_{2k+1}=0 \iff y_{\text{o}}^{2k+1}(x)=y_{\text{o}}^{2k_0+1}(x), \quad \forall k>k_0
+    \end{equation*}
+    \end{itemize} 
+    
+    There is the possibility of expressing the solution in Eq.\eqref{eq:solution} in a more compact form, combining the two solutions $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$. They are both a polynomial of maximum degree $n$, assuming $n \in \mathbb{N}$. In the case where $n$ is even, the polynomial solution
+    \begin{equation*}
+    \lim_{K\to \infty} y_\text{e}^K(x)
+    \end{equation*}
+    will be a finite sum. If instead $n$ is odd, will be 
+    \begin{equation*}
+    \lim_{K\to \infty} y_\text{o}^K(x)
+    \end{equation*}
+    to be a finite sum. 
+    
+    Depending on the coefficient we start with, $a_1$ or $a_0$, we will obtain the odd or even polynomial respectively. Starting with the last coefficient $a_n$ and, recursively, calculating all the others in descending order, we can express the two parts $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$ with a single sum. Hence, because we start with the last coefficient, the choice concerning $a_1$ and $a_0$ will be at the end of the sum, and not at the beginning. To compact Eq.\eqref{eq:solution}, Eq.\eqref{eq:recursion} can be reconsidered to calculate the coefficient $a_{k-2}$, using $a_k$
+    \begin{equation*}
+    a_{k-2} = -\frac{(k+2)(k+1)}{(k-n)(k+n+1)}a_k
+    \end{equation*}
+    Now the game is to find a pattern, as before. Remember that $n$ is a fixed parameter of Eq.\eqref{eq:legendre}.  
+    \begin{align*}
+    a_{n-2} &= -\frac{n(n-1)}{2(2n-1)}a_n, \\
+    a_{n-4} &= -\frac{(n-2)(n-3)}{4(2n-3)}a_{n-2} \\
+    &= -\frac{(n-2)(n-3)}{4(2n-3)}-\frac{n(n-1)}{2(2n-1)}a_n.
+    \end{align*}
+    In general 
+    \begin{equation}\label{eq:general_recursion}
+    a_{n-2k} = (-1)^k \frac{n(n-1)(n-2)(n-3) \hdots (n-2k+1)}{2\cdot4\hdots 2k(2n-1)(2n-3)\hdots(2n-2k+1)}a_n
+    \end{equation}
+    The whole solution can now be written as
+    \begin{align}
+    y(x) &= a_n x^n + a_{n-2} x^{n-2} + a_{n-4} x^{n-4} + a_{n-6} x^{n-6} + \hdots + \begin{cases} 
+    a_1 x, \quad &\text{if } n \text{ odd} \\ 
+    a_0, \quad  &\text{if } n \text{ even} 
+    \end{cases} \nonumber \\
+    &= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} a_{n-2k}x^{n-2k} \label{eq:solution_2}
+    \end{align}
+    By considering
+    \begin{align}
+    (2n-1)(2n-3)\hdots (2n-2k+1)&=\frac{2n(2n-1)(2n-2)(2n-3)\hdots(2n-2k+1)}
+    {2n(2n-2)(2n-4)(2n-6)\hdots(2n-2k+2)} \nonumber \\ 
+    &=\frac{\frac{(2n)!}{(2n-2k)!}}
+    {2^kn(n-1)(n-2)(n-3)\hdots(n-k+1)} \nonumber \\
+    &=\frac{\frac{(2n)!}{(2n-2k)!}}
+    {2^k\frac{n!}{(n-k)!}}=\frac{(n-k)!(2n)!}{n!(2n-2k)!2^k} \label{eq:1_sub_recursion}, \\
+    2 \cdot 4  \hdots 2k &= 2^r 1\cdot2 \hdots r = 2^r r!\label{eq:2_sub_recursion}, \\
+    n(n-1)(n-2)(n-3) \hdots (n-2k+1) &= \frac{n!}{(n-2k)!}\label{eq:3_sub_recursion}.
+    \end{align}
+    Eq.\eqref{eq:solution_2} can be rewritten as
+    \begin{equation}\label{eq:solution_3}
+    y(x)=a_n \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{n!^2(2n-2k)!}{k!(n-2k)!(n-k)!(2n)!}  x^{n-2k}.
+    \end{equation}
+    Eq.\eqref{eq:solution_3} is defined for any $a_n$. By letting $a_n$ be declared as
+    \begin{equation*}
+    a_{n} := \frac{(2n)!}{2^n n!^2},
+    \end{equation*}
+    the so called \emph{Legendre polynomial} emerges
+    \begin{equation}\label{eq:leg_poly}
+    P_n(x):=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k} 
+    \end{equation}
+\end{proof}
+
+
+\begin{lemma}
+  If $Z_n(z)$ is a solution of the Legendre equation \eqref{kugel:eqn:legendre},
+  then
+  \begin{equation*}
+    P^m_n(z) = (1 - z^2)^{m/2} \frac{d^m}{dz^m}Z_n(z)
+  \end{equation*}
+  solves the associated Legendre equation \eqref{kugel:eqn:associated-legendre}.
+\end{lemma}
+% \begin{proof} [TODO: modificare la $m$ (è già usata come costante di separazione) o forse è giusta (?)]
+\begin{proof}
+    To begin, we can start by differentiating $m$ times Eq.\eqref{kugel:eq:leg_eq} (which is staisfied by $y(x)$), obtaining
+    \begin{equation}\label{eq:lagrange_mderiv}
+    \frac{d^m}{dx^m}\left[ (1-x^2)\frac{d^2y}{dx^2} \right] -2 \frac{d^m}{dx^m}\left[ x\frac{dy}{dx} \right] + n(n+1)\frac{d^m}{dx^m}y=0.
+    \end{equation}
+    \emph{Leibniz's theorem} says, that if we want to differentiate $m$ times a multiplication of two functions, we can use the binomial coefficients to build up a sum. This allows us to be more compact, obtaining 
+    \begin{equation}\label{eq:leibniz}
+    \frac{d^m}{dx^m}[u(x)v(x)] = \sum_{i=0}^m \binom{n}{i} \frac{d^{m-i}u}{dx^{m-1}} \frac{d^{i}v}{dx^i}.
+    \end{equation}
+    Using Eq.\eqref{eq:leibniz} in Eq.\eqref{eq:lagrange_mderiv}, we have
+    \begin{align}
+    (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} &+ m \frac{d}{dx}(1-x^2)\frac{d^{m+1}y}{dx^{m+1}} + \frac{m(m-1)}{2}\frac{d^{2}}{dx^{2}}(1-x^2)\frac{d^{m}y}{dx^{m}} + n(n+1)\frac{d^m{}y}{dx^{m}} \nonumber \\
+    &-2\left(x\frac{d^{m+1}y}{dx^{m+1}} + m\frac{d}{dx}x\frac{d^{m}y}{dx^{m}} \right) \nonumber \\
+    &= (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} -2x(m+1)\frac{d^{m+1}y}{dx^{m+1}}+(n(n+1)-m(m-1)-2m)\frac{d^{m}y}{dx^{m}}=0. \label{eq:aux_3}
+    \end{align}
+    To make the notation easier to follow, a new function can be defined
+    \begin{equation*}
+    \frac{d^{m}y}{dx^{m}} := y_m.
+    \end{equation*}
+    Eq.\eqref{eq:aux_3} now becomes
+    \begin{equation}\label{eq:1st_subs}
+    (1-x^2)\frac{d^{2}y_m}{dx^{2}} -2x(m+1)\frac{dy_m}{dx}+(n(n+1)-m(m+1))y_m=0
+    \end{equation}
+    A second function can be further defined as
+    \begin{equation*}
+    (1-x^2)^{\frac{m}{2}}\frac{d^{m}y}{dx^{m}} = (1-x^2)^{\frac{m}{2}}y_m := \hat{y}_m,
+    \end{equation*}
+    allowing to write Eq.\eqref{eq:1st_subs} as
+    \begin{equation}\label{eq:2st_subs}
+    (1-x^2)\frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] -2(m+1)x\frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] + (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0.
+    \end{equation}
+    The goal now is to compute the two terms 
+    \begin{align*}
+    \frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &=  \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\
+    &+ m\left( \frac{d\hat{y}_m}{dx} x (1-x^2)^{-\frac{m}{2}-1} + \hat{y}_m (1-x^2)^{-\frac{m}{2}-1} - \hat{y}_m x (-\frac{m}{2}-1)(1-x^2)^{-\frac{m}{2}} 2x\right) \\
+    &= \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1}\\
+    &+ m\hat{y}_m  (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2}
+    \end{align*}
+    and
+    \begin{align*}
+    \frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_m\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\
+    &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x,
+    \end{align*}
+    to use them in Eq.\eqref{eq:2st_subs}, obtaining
+    \begin{align*}
+    (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1} \\ 
+    &+ m\hat{y}_m  (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2}\biggr] \\
+    &-2(m+1)x\left[  \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x \right] \\
+    &+ (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0.\\
+    \end{align*}
+    We can now divide by $(1-x^2)^{-\frac{m}{2}}$, obtaining
+    \begin{align*}
+    (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-1} + m\hat{y}_m  (1-x^2)^{-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-2}\biggr] \\
+    &-2(m+1)x\left[  \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\
+    &= \frac{d^2\hat{y}_m}{dx^2} + \frac{d\hat{y}_m}{dx}mx + m\frac{d\hat{y}_m}{dx}x + m\hat{y}_m + m\hat{y}_m x^2(m+2)(1-x^2)^{-1} \\
+    &-2(m+1)x\left[  \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\
+    \end{align*}
+    and collecting some terms
+    \begin{equation*}
+    (1-x^2)\frac{d^2\hat{y}_m}{dx^2} - 2x\frac{d\hat{y}_m}{dx} + \left( -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1)\right)\hat{y}_m=0.
+    \end{equation*}
+    Showing that 
+    \begin{align*}
+    -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1) &= n(n+1)- m^2 -x^2 \frac{m^2}{1-x^2} \\
+    &= n(n+1)- \frac{m}{1-x^2}
+    \end{align*}
+    implies $\hat{y}_m(x)$ being a solution of Eq.\eqref{kugel:eq:associated_leg_eq}
+\end{proof}
diff --git a/buch/papers/kugel/references.bib b/buch/papers/kugel/references.bib
index b74c5cd..e5d6452 100644
--- a/buch/papers/kugel/references.bib
+++ b/buch/papers/kugel/references.bib
@@ -192,4 +192,15 @@ Created by Henry Reich},
 	urldate = {2022-08-01},
 	date = {2022},
 	file = {Metric Spaces\: Completeness:/Users/npross/Zotero/storage/5JYEE8NF/completeness.html:text/html},
+}
+
+@book{bell_special_2004,
+	location = {Mineola, {NY}},
+	title = {Special functions for scientists and engineers},
+	isbn = {978-0-486-43521-3},
+	series = {Dover books on mathematics},
+	pagetotal = {247},
+	publisher = {Dover Publ},
+	author = {Bell, William Wallace},
+	date = {2004},
 }
\ No newline at end of file
diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex
index 6b23ce5..2ded50b 100644
--- a/buch/papers/kugel/spherical-harmonics.tex
+++ b/buch/papers/kugel/spherical-harmonics.tex
@@ -1,13 +1,410 @@
-% vim:ts=2 sw=2 et spell:
+% vim:ts=2 sw=2 et spell tw=80:
 
-\section{Spherical Harmonics}
+\section{Construction of the Spherical Harmonics}
 
-\subsection{Eigenvalue Problem in Spherical Coordinates}
+\kugeltodo{Review text, or rewrite if preliminaries becomes an addendum}
+
+We finally arrived at the main section, which gives our chapter its name. The
+idea is to discuss spherical harmonics, their mathematical derivation and some
+of their properties and applications.
+
+The subsection \ref{} \kugeltodo{Fix references} will be devoted to the
+Eigenvalue problem of the Laplace operator. Through the latter we will derive
+the set of Eigenfunctions that obey the equation presented in \ref{}
+\kugeltodo{reference to eigenvalue equation}, which will be defined as
+\emph{Spherical Harmonics}. In fact, this subsection will present their
+mathematical derivation.
+
+In the subsection \ref{}, on the other hand, some interesting properties
+related to them will be discussed. Some of these will come back to help us
+understand in more detail why they are useful in various real-world
+applications, which will be presented in the section \ref{}.
+
+One specific property will be studied in more detail in the subsection \ref{},
+namely the recursive property.  The last subsection is devoted to one of the
+most beautiful applications (In our humble opinion), namely the derivation of a
+Fourier-style series expansion but defined on the sphere instead of a plane.
+More importantly, this subsection will allow us to connect all the dots we have
+created with the previous sections, concluding that Fourier is just a specific
+case of the application of the concept of orthogonality. Our hope is that after
+reading this section you will appreciate the beauty and power of generalization
+that mathematics offers us.
+
+\subsection{Eigenvalue Problem}
+\label{kugel:sec:construction:eigenvalue}
+
+\begin{figure}
+  \centering
+  \includegraphics{papers/kugel/figures/tikz/spherical-coordinates}
+  \caption{
+    Spherical coordinate system. Space is described with the free variables $r
+    \in \mathbb{R}_0^+$, $\vartheta \in [0; \pi]$ and $\varphi \in [0; 2\pi)$.
+    \label{kugel:fig:spherical-coordinates}
+  }
+\end{figure}
+
+From Section \ref{buch:pde:section:kugel}, we know that the spherical Laplacian
+in the spherical coordinate system (shown in Figure
+\ref{kugel:fig:spherical-coordinates}) is is defined as
+\begin{equation*}
+    \sphlaplacian :=
+      \frac{1}{r^2} \frac{\partial}{\partial r} \left(
+        r^2 \frac{\partial}{\partial r}
+      \right)
+      + \frac{1}{r^2} \left[
+          \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left(
+            \sin\vartheta \frac{\partial}{\partial\vartheta}
+          \right)
+        + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}
+      \right].
+\end{equation*}
+But we will not consider this algebraic monstrosity in its entirety. As the
+title suggests, we will only care about the \emph{surface} of the sphere.  This
+is for many reasons, but mainly to simplify reduce the already broad scope of
+this text. Concretely, we will always work on the unit sphere, which just means
+that we set $r = 1$ and keep only $\vartheta$ and $\varphi$ as free variables.
+Now, since the variable $r$ became a constant, we can leave out all derivatives
+with respect to $r$ and substitute all $r$'s with 1's to obtain a new operator
+that deserves its own name.
+
+\begin{definition}[Surface spherical Laplacian]
+  \label{kugel:def:surface-laplacian}
+  The operator
+  \begin{equation*}
+      \surflaplacian :=
+        \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left(
+          \sin\vartheta \frac{\partial}{\partial\vartheta}
+        \right)
+        + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2},
+  \end{equation*}
+  is called the surface spherical Laplacian.
+\end{definition}
+
+In the definition, the subscript ``$\partial S$'' was used to emphasize the
+fact that we are on the spherical surface, which can be understood as being the
+boundary of the sphere. But what does it actually do? To get an intuition,
+first of all, notice the fact that $\surflaplacian$ have second derivatives,
+which means that this a measure of \emph{curvature}; But curvature of what? To
+get an even stronger intuition we will go into geometry, were curvature can be
+grasped very well visually. Consider figure \ref{kugel:fig:curvature} where the
+curvature is shown using colors. First we have the curvature of a curve in 1D,
+then the curvature of a surface (2D), and finally the curvature of a function on
+the surface of the unit sphere.
+
+\begin{figure}
+  \centering
+  \includegraphics[width=.3\linewidth]{papers/kugel/figures/tikz/curvature-1d}
+  \hskip 5mm
+  \includegraphics[width=.3\linewidth]{papers/kugel/figures/povray/curvature}
+  \hskip 5mm
+  \includegraphics[width=.3\linewidth]{papers/kugel/figures/povray/spherecurve}
+  \caption{
+    \kugeltodo{Fix alignment / size, add caption. Would be nice to match colors.}
+    \label{kugel:fig:curvature}
+  }
+\end{figure}
+
+Now that we have defined an operator, we can go and study its eigenfunctions,
+which means that we would like to find the functions $f(\vartheta, \varphi)$
+that satisfy the equation
+\begin{equation} \label{kuvel:eqn:eigen}
+    \surflaplacian f = -\lambda f.
+\end{equation}
+Perhaps it may not be obvious at first glance, but we are in fact dealing with a
+partial differential equation (PDE) \kugeltodo{Boundary conditions?}. If we
+unpack the notation of the operator $\nabla^2_{\partial S}$ according to
+definition
+\ref{kugel:def:surface-laplacian}, we get:
+\begin{equation} \label{kugel:eqn:eigen-pde}
+    \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left(
+      \sin\vartheta \frac{\partial f}{\partial\vartheta}
+    \right)
+    + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 f}{\partial\varphi^2}
+    + \lambda f = 0.
+\end{equation}
+Since all functions satisfying \eqref{kugel:eqn:eigen-pde} are the
+\emph{eigenfunctions} of $\surflaplacian$, our new goal is to solve this PDE.
+The task may seem very difficult but we can simplify it with a well-known
+technique: \emph{the separation Ansatz}. It consists in assuming that the
+function $f(\vartheta, \varphi)$ can be factorized in the following form:
+\begin{equation}
+    f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi). 
+\end{equation}
+In other words, we are saying that the effect of the two independent variables
+can be described using the multiplication of two functions that describe their
+effect separately. This separation process was already presented in section
+\ref{buch:pde:section:kugel}, but we will briefly rehearse it here for
+convenience. If we substitute this assumption in
+\eqref{kugel:eqn:eigen-pde}, we have:
+\begin{equation*}
+    \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left(
+      \sin\vartheta \frac{\partial  \Theta(\vartheta)}{\partial\vartheta}
+    \right) \Phi(\varphi)
+    + \frac{1}{\sin^2 \vartheta}
+      \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2}
+      \Theta(\vartheta)
+    + \lambda \Theta(\vartheta)\Phi(\varphi) = 0.
+\end{equation*}
+Dividing by $\Theta(\vartheta)\Phi(\varphi)$ and introducing an auxiliary
+variable $m^2$, the separation constant, yields:
+\begin{equation*}
+  \frac{1}{\Theta(\vartheta)}\sin \vartheta \frac{d}{d \vartheta} \left(
+    \sin \vartheta \frac{d \Theta}{d \vartheta}
+  \right)
+  + \lambda \sin^2 \vartheta
+  = -\frac{1}{\Phi(\varphi)} \frac{d^2\Phi(\varphi)}{d\varphi^2}
+  = m^2,
+\end{equation*}
+which is equivalent to the following system of 2 first order differential
+equations (ODEs):
+\begin{subequations}
+  \begin{gather}
+    \frac{d^2\Phi(\varphi)}{d\varphi^2} = -m^2 \Phi(\varphi),
+      \label{kugel:eqn:ode-phi} \\ 
+    \sin \vartheta \frac{d}{d \vartheta} \left(
+      \sin \vartheta \frac{d \Theta}{d \vartheta}
+    \right)
+    + \left( \lambda - \frac{m^2}{\sin^2 \vartheta} \right)
+      \Theta(\vartheta) = 0
+      \label{kugel:eqn:ode-theta}.
+  \end{gather}
+\end{subequations}
+The solution of \eqref{kugel:eqn:ode-phi} is easy to find: The complex
+exponential is obviously the function we are looking for. So we can directly
+write the solutions
+\begin{equation} \label{kugel:eqn:ode-phi-sol}
+    \Phi(\varphi) = e^{i m \varphi}, \quad m \in \mathbb{Z}.
+\end{equation}
+The restriction that the separation constant $m$ needs to be an integer arises
+from the fact that we require a $2\pi$-periodicity in $\varphi$ since the
+coordinate systems requires that $\Phi(\varphi + 2\pi) = \Phi(\varphi)$.
+Unfortunately, solving \eqref{kugel:eqn:ode-theta} is as straightforward,
+actually, it is quite difficult, and the process is so involved that it will
+require a dedicated section of its own.
+
+\subsection{Legendre Functions}
+
+\begin{figure}
+  \centering
+  \kugelplaceholderfig{.8\textwidth}{5cm}
+  \caption{
+    \kugeltodo{Why $z = \cos \vartheta$.}
+  }
+\end{figure}
+
+To solve \eqref{kugel:eqn:ode-theta} we start with the substitution $z = \cos
+\vartheta$ \kugeltodo{Explain geometric origin with picture}. The operator
+$\frac{d}{d \vartheta}$ becomes
+\begin{equation*}
+    \frac{d}{d \vartheta}
+    = \frac{dz}{d \vartheta}\frac{d}{dz}
+    = -\sin \vartheta \frac{d}{dz}
+    = -\sqrt{1-z^2} \frac{d}{dz},
+\end{equation*} 
+since $\sin \vartheta = \sqrt{1 - \cos^2 \vartheta} = \sqrt{1 - z^2}$, and
+then \eqref{kugel:eqn:ode-theta} becomes 
+\begin{align*}
+  \frac{-\sqrt{1-z^2}}{\sqrt{1-z^2}} \frac{d}{dz} \left[
+    \left(\sqrt{1-z^2}\right) \left(-\sqrt{1-z^2}\right) \frac{d \Theta}{dz}
+  \right]
+  + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0,
+  \\
+  \frac{d}{dz} \left[ (1-z^2) \frac{d \Theta}{dz} \right]
+  + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0,
+  \\
+  (1-z^2)\frac{d^2 \Theta}{dz} - 2z\frac{d \Theta}{dz}
+  + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0.
+\end{align*}
+By making two final cosmetic substitutions, namely $Z(z) = \Theta(\cos^{-1}z)$
+and $\lambda = n(n+1)$, we obtain what is known in the literature as the
+\emph{associated Legendre equation of order $m$}:
+\nocite{olver_introduction_2013}
+\begin{equation} \label{kugel:eqn:associated-legendre}
+  (1 - z^2)\frac{d^2 Z}{dz}
+  - 2z\frac{d Z}{dz}
+  + \left( n(n + 1) - \frac{m^2}{1 - z^2} \right) Z(z) = 0,
+  \quad
+  z \in [-1; 1], m \in \mathbb{Z}.
+\end{equation}
+
+Our new goal has therefore become to solve
+\eqref{kugel:eqn:associated-legendre}, since if we find a solution for $Z(z)$ we
+can perform the substitution backwards and get back to our eigenvalue problem.
+However, the associated Legendre equation is not any easier, so to attack the
+problem we will look for the solutions in the easier special case when $m = 0$.
+This reduces the problem because it removes the double pole, which is always
+tricky to deal with. In fact, the reduced problem when $m = 0$ is known as the
+\emph{Legendre equation}:
+\begin{equation} \label{kugel:eqn:legendre}
+  (1 - z^2)\frac{d^2 Z}{dz}
+  - 2z\frac{d Z}{dz}
+  + n(n + 1) Z(z) = 0,
+  \quad
+  z \in [-1; 1].
+\end{equation}
+
+The Legendre equation is a second order differential equation, and therefore it
+has 2 independent solutions, which are known as \emph{Legendre functions} of the
+first and second kind. For the scope of this text we will only derive a special
+case of the former that is known known as the \emph{Legendre polynomials}, since
+we only need a solution between $-1$ and $1$.
+
+\begin{lemma}[Legendre polynomials]
+  \label{kugel:lem:legendre-poly}
+  The polynomial function
+  \[
+    P_n(z) = \sum^{\lfloor n/2 \rfloor}_{k=0}
+      \frac{(-1)^k}{2^n s^k!} \frac{(2n - 2k)!}{(n - k)! (n-2k)!} z^{n - 2k}
+  \]
+  is the only finite solution of the Legendre equation
+  \eqref{kugel:eqn:legendre} when $n \in \mathbb{Z}$ and $z \in [-1; 1]$.
+\end{lemma}
+\begin{proof}
+  This results is derived in section \ref{kugel:sec:proofs:legendre}.
+\end{proof}
+
+Since the Legendre \emph{polynomials} are indeed polynomials, they can also be
+expressed using the hypergeometric functions described in section
+\ref{buch:rekursion:section:hypergeometrische-funktion}, so in fact
+\begin{equation}
+  P_n(z) = {}_2F_1 \left( \begin{matrix}
+    n + 1, & -n \\ \multicolumn{2}{c}{1}
+  \end{matrix} ; \frac{1 - z}{2} \right).
+\end{equation}
+Further, there are a few more interesting but not very relevant forms to write
+$P_n(z)$ such as \emph{Rodrigues' formula} and \emph{Laplace's integral
+representation} which are
+\begin{equation*}
+  P_n(z) = \frac{1}{2^n} \frac{d^n}{dz^n} (x^2 - 1)^n,
+  \qquad \text{and} \qquad
+  P_n(z) = \frac{1}{\pi} \int_0^\pi \left(
+    z + \cos\vartheta \sqrt{z^2 - 1}
+  \right) \, d\vartheta
+\end{equation*}
+respectively, both of which we will not prove (see chapter 3 of
+\cite{bell_special_2004} for a proof). Now that we have a solution for the
+Legendre equation, we can make use of the following lemma patch the solutions
+such that they also become solutions of the associated Legendre equation
+\eqref{kugel:eqn:associated-legendre}.
+
+\begin{lemma} \label{kugel:lem:extend-legendre}
+  If $Z_n(z)$ is a solution of the Legendre equation \eqref{kugel:eqn:legendre},
+  then
+  \begin{equation*}
+    Z^m_n(z) = (1 - z^2)^{m/2} \frac{d^m}{dz^m}Z_n(z)
+  \end{equation*}
+  solves the associated Legendre equation \eqref{kugel:eqn:associated-legendre}.
+  \nocite{bell_special_2004}
+\end{lemma}
+\begin{proof}
+  See section \ref{kugel:sec:proofs:legendre}.
+\end{proof}
+
+What is happening in lemma \ref{kugel:lem:extend-legendre}, is that we are
+essentially inserting a square root function in the solution in order to be able
+to reach the parts of the domain near the poles at $\pm 1$ of the associated
+Legendre equation, which is not possible only using power series
+\kugeltodo{Reference book theory on extended power series method.}. Now, since
+we have a solution in our domain, namely $P_n(z)$, we can insert it in the lemma 
+obtain the \emph{associated Legendre functions}.
+
+\begin{definition}[Ferrers or associated Legendre functions]
+  \label{kugel:def:ferrers-functions}
+  The functions
+  \begin{equation}
+    P^m_n (z) = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m}}{dz^{m}} P_n(z)
+      = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m+n}}{dz^{m+n}}(1-z^2)^n 
+  \end{equation}
+  are known as Ferrers or associated Legendre functions.
+\end{definition}
+
+\kugeltodo{Discuss $|m| \leq n$.}
+
+\if 0
+The constraint $|m|<n$, can be justified by considering Eq.\eqref{kugel:eq:associated_leg_func}, in which the derivative of degree $m+n$ is present. A derivative to be well defined must have an order that is greater than zero. Furthermore, it can be seen that this derivative is applied on a polynomial of degree $2n$. As is known from Calculus 1, if you derive a polynomial of degree $2n$ more than $2n$ times, you get zero, which is a trivial solution in which we are not interested.\newline
+We can thus summarize these two conditions by writing:
+\begin{equation*}
+    \begin{rcases}
+        m+n \leq 2n &\implies m \leq n \\
+        m+n \geq 0  &\implies  m \geq -n
+    \end{rcases} |m| \leq n.
+\end{equation*}
+The set of functions in Eq.\eqref{kugel:eq:sph_harm_0} is named \emph{Spherical Harmonics}, which are the eigenfunctions of the Laplace operator on the \emph{spherical surface domain}, which is exactly what we were looking for at the beginning of this section.
+\fi
+
+\subsection{Spherical Harmonics}
+
+Finally, we can go back to solving our boundary value problem we started in
+section \ref{kugel:sec:construction:eigenvalue}. We had left off in the middle
+of the separation, were we had used the Ansatz $f(\vartheta, \varphi) =
+\Theta(\vartheta) \Phi(\varphi)$ to find that $\Phi(\varphi) = e^{im\varphi}$,
+and we were solving for $\Theta(\vartheta)$.  As you may recall, previously we
+performed the substitution $z = \cos \vartheta$. Now we can finally to bring back the
+solution to the associated Legendre equation $P^m_n(z)$ into the $\vartheta$
+domain and combine it with $\Phi(\varphi)$ to get the full result:
+\begin{equation*}
+    f(\vartheta, \varphi)
+      = \Theta(\vartheta)\Phi(\varphi)
+      = P^m_n (\cos \vartheta) e^{im\varphi}.
+\end{equation*}
+This family of functions, which recall are the solutions of the eigenvalue
+problem of the surface spherical Laplacian, are the long anticipated
+\emph{complex spherical harmonics}, and they are usually denoted with
+$Y^m_n(\vartheta, \varphi)$.
+
+\begin{definition}[Spherical harmonics]
+  \label{kugel:def:spherical-harmonics}
+  The functions
+  \begin{equation*}
+    Y_{m,n}(\vartheta, \varphi) = P^m_n(\cos \vartheta) e^{im\varphi},
+  \end{equation*}
+  where $m, n \in \mathbb{Z}$ and $|m| < n$ are called spherical harmonics.
+\end{definition}
+
+\begin{figure}
+  \centering
+  \kugelplaceholderfig{\textwidth}{.8\paperheight}
+  \caption{
+    \kugeltodo{Big picture with the first few spherical harmonics.}
+  }
+\end{figure}
+
+\subsection{Normalization}
+
+\kugeltodo{Discuss various normalizations.}
+
+\if 0
+As explained in the chapter \ref{}, the concept of orthogonality is very important and at the practical level it is very useful, because it allows us to develop very powerful techniques at the mathematical level.\newline 
+Throughout this book we have been confronted with the Sturm-Liouville theory (see chapter \ref{}). The latter, among other things, carries with it the concept of orthogonality. Indeed, if we consider the solutions of the Sturm-Liouville equation, which can be expressed in this form
+\begin{equation}\label{kugel:eq:sturm_liouville}
+    \mathcal{S}f := \frac{d}{dx}\left[p(x)\frac{df}{dx}\right]+q(x)f(x)
+\end{equation}
+possiamo dire che formano una base ortogonale.\newline
+Adesso possiamo dare un occhiata alle due equazioni che abbiamo ottenuto tramite la Separation Ansatz (Eqs.\eqref{kugel:eq:associated_leg_eq}\eqref{kugel:eq:ODE_1}), le quali possono essere riscritte come:
+\begin{align*}
+    \frac{d}{dx} \left[ (1-x^2) \cdot \frac{dP_{m,n}}{dx} \right] &+ \left(n(n+1)-\frac{m}{1-x^2} \right) \cdot P_{m,n}(x) = 0, \\
+    \frac{d}{d\varphi} \left[ 1 \cdot \frac{ d\Phi }{d\varphi} \right] &+ 1 \cdot \Phi(\varphi) = 0. 
+\end{align*}
+Si può concludere in modo diretto che sono due casi dell'equazione di Sturm-Liouville. Questo significa che le loro soluzioni sono ortogonali sotto l'inner product con weight function $w(x)=1$, dunque:
+\begin{align}
+\int_{0}^{2\pi} \Phi_m(\varphi)\Phi_m'(\varphi) d\varphi &= \delta_{m'm}, \nonumber \\
+\int_{-1}^1 P_{m,m'}(x)P_{n,n'}(x) dx &= \delta_{m'm}\delta_{n'n}. \label{kugel:eq:orthogonality_associated_func}
+\end{align}
+Inoltre, possiamo provare l'ortogonalità di $\Theta(\vartheta)$ utilizzando \eqref{kugel:eq:orthogonality_associated_func}:
+\begin{align}
+    x
+\end{align}
+Ora, visto che la soluzione dell'eigenfunction problem è formata dalla moltiplicazione di $\Phi_m(\varphi)$ e $P_{m,n}(x)$
+\fi
 
 \subsection{Properties}
 
 \subsection{Recurrence Relations}
 
-\section{Series Expansions in \(C(S^2)\)}
+\section{Series Expansions in $C(S^2)$}
 
-\nocite{olver_introduction_2013}
+\subsection{Orthogonality of $P_n$, $P^m_n$ and $Y^m_n$}
+
+\subsection{Series Expansion}
+
+\subsection{Fourier on $S^2$}
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..964b348
--- /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..42cae0d 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..f09edfb 100644
--- a/buch/papers/lambertw/Bilder/Strategie.py
+++ b/buch/papers/lambertw/Bilder/Strategie.py
@@ -9,6 +9,9 @@ import pylatex
 import numpy as np
 import matplotlib.pyplot as plt
 
+
+
+
 N = np.array([0, 0])
 V = np.array([1, 4])
 Z = np.array([5, 5])
@@ -35,7 +38,9 @@ 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.tick_params(labelsize=15)
+plt.xticks(ticks=range(0, 7))
+plt.yticks(ticks=range(0, 7))
 ax.text(2.5, 4.5, "Visierlinie", size=20, rotation=10)
 
 plt.rcParams.update({
@@ -44,9 +49,10 @@ 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')
+ax.set_xlabel(r"$x$", size=20)
+ax.set_ylabel(r"$y$", size=20)
 
 
diff --git a/buch/papers/lambertw/Bilder/Strategie.svg b/buch/papers/lambertw/Bilder/Strategie.svg
deleted file mode 100644
index 30f9f22..0000000
--- a/buch/papers/lambertw/Bilder/Strategie.svg
+++ /dev/null
@@ -1,790 +0,0 @@
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-   <cc:Work>
-    <dc:type rdf:resource="http://purl.org/dc/dcmitype/StillImage"/>
-    <dc:date>2022-07-29T16:52:06.315252</dc:date>
-    <dc:format>image/svg+xml</dc:format>
-    <dc:creator>
-     <cc:Agent>
-      <dc:title>Matplotlib v3.3.2, https://matplotlib.org/</dc:title>
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diff --git a/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png b/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png
index 90758cd..dc4720a 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..73b322c
--- /dev/null
+++ b/buch/papers/lambertw/Bilder/lambertAbstandBauchgefühl.py
@@ -0,0 +1,62 @@
+# -*- 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.grid(True)
+plt.tick_params(labelsize=15)
+#plt.xticks(ticks=range(0, 7))
+#plt.yticks(ticks=range(0, 7))
+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')
+plt.ylabel(r"$y$", size=20)
+plt.xlabel(r"$x$", size=20)
\ No newline at end of file
diff --git a/buch/papers/lambertw/Bilder/pursuerDGL.ggb b/buch/papers/lambertw/Bilder/pursuerDGL.ggb
deleted file mode 100644
index 3fb3a78..0000000
--- a/buch/papers/lambertw/Bilder/pursuerDGL.ggb
+++ /dev/null
diff --git a/buch/papers/lambertw/Bilder/pursuerDGL.svg b/buch/papers/lambertw/Bilder/pursuerDGL.svg
deleted file mode 100644
index d91e5e1..0000000
--- a/buch/papers/lambertw/Bilder/pursuerDGL.svg
+++ /dev/null
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255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="463" y="1288">0.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="463" y="1288">0.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="515" y="1288">0.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="515" y="1288">0.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="515" y="1288">0.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="566" y="1288">0.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="566" y="1288">0.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="566" y="1288">0.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="618" y="1288">0.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="618" y="1288">0.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="618" y="1288">0.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="670" y="1288">0.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="670" y="1288">0.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="670" y="1288">0.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="721" y="1288">0.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="721" y="1288">0.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="721" y="1288">0.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="773" y="1288">0.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="773" y="1288">0.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="773" y="1288">0.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="825" y="1288">0.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="825" y="1288">0.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="825" y="1288">0.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="876" y="1288">0.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="876" y="1288">0.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="876" y="1288">0.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="934" y="1288">1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="934" y="1288">1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="934" y="1288">1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="980" y="1288">1.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="980" y="1288">1.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="980" y="1288">1.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1032" y="1288">1.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1032" y="1288">1.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1032" y="1288">1.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1083" y="1288">1.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1083" y="1288">1.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1083" y="1288">1.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1135" y="1288">1.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1135" y="1288">1.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1135" y="1288">1.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1187" y="1288">1.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1187" y="1288">1.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1187" y="1288">1.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1238" y="1288">1.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1238" y="1288">1.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1238" y="1288">1.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1290" y="1288">1.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1290" y="1288">1.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1290" y="1288">1.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1342" y="1288">1.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1342" y="1288">1.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1342" y="1288">1.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1393" y="1288">1.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1393" y="1288">1.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1393" y="1288">1.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1451" y="1288">2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1451" y="1288">2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1451" y="1288">2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1497" y="1288">2.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1497" y="1288">2.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1497" y="1288">2.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1549" y="1288">2.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1549" y="1288">2.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1549" y="1288">2.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1600" y="1288">2.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1600" y="1288">2.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1600" y="1288">2.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1652" y="1288">2.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1652" y="1288">2.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1652" y="1288">2.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1704" y="1288">2.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1704" y="1288">2.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1704" y="1288">2.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1755" y="1288">2.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1755" y="1288">2.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1755" y="1288">2.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1807" y="1288">2.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1807" y="1288">2.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1807" y="1288">2.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1859" y="1288">2.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1859" y="1288">2.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1859" y="1288">2.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1910" y="1288">2.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1910" y="1288">2.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1910" y="1288">2.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1968" y="1288">3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="1968" y="1288">3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="1968" y="1288">3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="2014" y="1288">3.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="2014" y="1288">3.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="2014" y="1288">3.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="2066" y="1288">3.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="2066" y="1288">3.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="2066" y="1288">3.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="2117" y="1288">3.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="2117" y="1288">3.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="2117" y="1288">3.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1226">0.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="1226">0.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1226">0.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1174">0.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="1174">0.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1174">0.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1122">0.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="1122">0.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1122">0.3</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1071">0.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="1071">0.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1071">0.4</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1019">0.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="1019">0.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="1019">0.5</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="967">0.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="967">0.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="967">0.6</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="916">0.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="916">0.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="916">0.7</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="864">0.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="864">0.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="864">0.8</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="812">0.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="812">0.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="812">0.9</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="405" y="760">1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="405" y="760">1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="405" y="760">1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="709">1.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="709">1.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="709">1.1</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="657">1.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="none" stroke="rgb(255, 255, 255)" stroke-linejoin="bevel" stroke-miterlimit="10" stroke-opacity="1" stroke-width="3" text-anchor="start" x="395" y="657">1.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" dominant-baseline="alphabetic" fill="rgb(0, 0, 0)" fill-opacity="1" stroke="none" text-anchor="start" x="395" y="657">1.2</text><text font-family="geogebra-sans-serif, sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" 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diff --git a/buch/papers/lambertw/teil0.tex b/buch/papers/lambertw/teil0.tex
index 8fa8f9b..baee9ea 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 +67,50 @@ 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}
-    \frac{z-v}{|z-v|}\cdot|\dot{v}|
-    =
     \dot{v}
+    \quad||\quad
+    z-v
+    \text{.}
 \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.
+Um den Richtungsvektor zu konstruieren kann der Einheitsvektor parallel zu $z-v$ um $|\dot{v}|$ gestreckt werden, was zu
+\begin{equation}
+    \dot{v}
+    =
+    |\dot{v}|\cdot (z-v)^\circ
+    =
+    |\dot{v}|\cdot\frac{z-v}{|z-v|}
+    \label{lambertw:richtungsvektor}
+\end{equation}
+führt.
 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
-    \\
+    \text{,}
+\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..c4b2d05 100644
--- a/buch/papers/lambertw/teil1.tex
+++ b/buch/papers/lambertw/teil1.tex
@@ -11,15 +11,16 @@ Sehr oft kommt es vor, dass bei Verfolgungsproblemen die Frage auftaucht, ob das
 Wenn zum Beispiel die Geschwindigkeit des Verfolgers kleiner ist als diejenige des Ziels, gibt es Anfangsbedingungen bei denen das Ziel nie erreicht wird.
 Im Anschluss dieser Frage stellt sich meist die nächste Frage, wie lange es dauert bis das Ziel erreicht wird.
 Diese beiden Fragen werden in diesem Kapitel behandelt und am Beispiel aus \ref{lambertw:section:teil4} betrachtet.
-Das Beispiel wird bei dieser Betrachtung noch etwas erweitert indem alle Punkte auf der gesamtem $xy$-Ebene als Startwerte zugelassen werden.
+Das Beispiel wird bei dieser Betrachtung noch etwas erweitert, indem alle Punkte auf der gesamtem $xy$-Ebene als Startwerte zugelassen werden.
 
 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
@@ -29,41 +30,42 @@ 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)} \\
+    x_0\cdot\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right) \right)} \text{,}\\
     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}\biggl(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\biggl(\left(\frac{x(t)}{x_0}\right)^2\biggr)-r_0+3y_0\biggr) \text{,}\\
     \chi
     &=
-    \frac{r_0+y_0}{r_0-y_0}, \quad
+    \frac{r_0+y_0}{r_0-y_0}\text{,} \quad
     \eta
     =
-    \left(\frac{x}{x_0}\right)^2,\quad
+    \left(\frac{x}{x_0}\right)^2 \quad\text{und}\quad
     r_0
     =
     \sqrt{x_0^2+y_0^2}
-\end{align*}
+\end{align}
 %
-Der Folger ist durch
+sind,
+beschrieben werden.
+Der Verfolger ist durch
 \begin{equation}
     v(t)
     =
     \left( \begin{array}{c} x(t) \\ y(t) \end{array} \right)
-    \text{.}
 \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)
@@ -74,35 +76,42 @@ 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}\biggl(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\biggl(\left(\frac{x(t)}{x_0}\right)^2\biggr)-r_0+3y_0\biggr)
+    \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.
-Somit kann nach den gestellten Bedingungen das Ziel nie erreicht werden.
+Dies nützt nicht viel, da unendlich viel Zeit vergehen müsste, damit ein Einholen möglich wäre.
+Somit kann unter den gestellten Bedingungen das Ziel nie erreicht werden.
 %
 %
 %
@@ -136,7 +145,7 @@ 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}
@@ -147,7 +156,7 @@ Dies kann veranschaulicht werden anhand
     1\text{.}
 \end{equation}
 %
-Da der $y$-Anteil der Geschwindigkeit des Ziels grösser-gleich der des Verfolgers ist, können die $y$-Koordinaten nie übereinstimmen.
+Da der $y$-Anteil der Geschwindigkeit des Ziels mindestens so gross wie die des Verfolgers ist, können die $y$-Koordinaten nie übereinstimmen.
 %
 \subsection{Anfangsbedingung auf positiven $y$-Achse}
 Wenn der Verfolger auf der positiven $y$-Achse startet, befindet er sich direkt auf der Fluchtgeraden des Ziels.
@@ -184,29 +193,153 @@ 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.
+
+Dieses Resultat ist aber eher akademischer Natur, weil der Verfolger und das Ziel als Punkt betrachtet wurden, während in der Realität nicht von Punkten sondern von Objekten mit einer räumlichen Ausdehnung gesprochen werden kann.
 Somit wird in einer nächsten Betrachtung untersucht, ob der Verfolger dem Ziel näher kommt als ein definierter Trefferradius.
 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.
-Durch quadrieren verschwindet die Wurzel des Betrages, womit
+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=0.7]{./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.
 
+$\dot{v}$ wird allgemein mit dem Winkel $\alpha \in[ 0, 2\pi)$ beschrieben, um alle unmittelbar benachbarten Punkte prüfen zu können.
+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 \\ 0 \end{array}\right)
+	\\
+	z
+	&=
+	\left(\begin{array}{c} 0 \\ t \end{array}\right)
+\end{align}
+beschrieben werden.
+$x_0$ ist der Abstand bei $t=0$, damit alle möglichen Fälle untersucht werden können.
+Da der Abstand allgemein
+\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)+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.
+Wenn für den Winkel zwischen den Richtungsvektoren $\alpha$ und die Eigenschaft $|\dot{z}|=|\dot{v}|$ verwendet wird entsteht
+\begin{equation}
+    \cos(\alpha)=1
+    \text{.}
+\end{equation}
+Jetzt ist klar, 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..36fb7e6 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'' beschrieben. 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. Aus diesen Bedingungen ergibt sich den ersten Quadranten als Bewegungsraum für \(V\). 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
@@ -40,39 +40,40 @@ Diese DGL haben wir bereits in Kapitel \ref{lambertw:subsection:Verfolger} defin
 
 \subsection{Differentialgleichung vereinfachen
 	\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.
+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 die Basis des Quadrates in \eqref{lambertw:eqAlgVerinfacht} 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.
@@ -117,7 +122,7 @@ Der Grund dafür ist, dass
 	\label{lambertw:eqQuotZeitAbleit}
 \end{equation}
 und somit kann der Quotient dieser zeitlichen Ableitungen in eine Ableitung nach \(x\) umgewandelt werden.
-Nach dem die Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eqVorKeineZeitAbleit} eingesetzt wird und vereinfacht wurde, entsteht die neue Gleichung
+Nachdem die Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eqVorKeineZeitAbleit} eingesetzt wurde, entsteht beim Vereinfachen die neue Gleichung
 \begin{equation}
 	x y^{\prime} + t - y
 	= 0.
@@ -125,31 +130,32 @@ Nach dem die Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eq
 \end{equation}
 
 \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 
+	\label{lambertw:subsubsection:VarTelimin}}
+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 Integrationsgrenzen vertauscht werden, was in einem Vorzeichenwechsel resultiert. 
 
-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)
 	&=
@@ -184,63 +191,65 @@ 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 exponentielle 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}}
 
+\definecolor{applegreen}{rgb}{0.55, 0.71, 0.0}
+
 \begin{figure}
 	\centering
 	\includegraphics{papers/lambertw/Bilder/VerfolgungskurveBsp.png}
-	\caption[Graph der Verfolgungskurve]{Graph der Verfolgungskurve wobei, ({\color{red}rot}) die Funktion \ensuremath{y(x)} ist, ({\color{darkgreen}grün}) der quadratische Teil und ({\color{blue}blau}) dem \ensuremath{\operatorname{ln}(x)}-Teil entspricht.
+	\caption[Graph der Verfolgungskurve]{Graph der Verfolgungskurve wobei, ({\color{red}rot}) die Funktion \ensuremath{y(x)} ist, ({\color{applegreen}grün}) der quadratische Teil und ({\color{blue}blau}) dem \ensuremath{\operatorname{ln}(x)}-Teil entspricht.
 	\label{lambertw:BildFunkLoes}
 	}
 \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{applegreen}{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 +264,17 @@ 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 +283,8 @@ 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 +294,18 @@ 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:
@@ -312,31 +324,32 @@ Nun sind wir soweit, dass wir eine \(y(x)\)-Beziehung für beliebige Anfangswert
 
 \subsection{Funktion nach der Zeit 
 	\label{lambertw:subsection:FunkNachT}}
-In diesem Abschnitt werden algebraischen Umformungen ein wenig detaillierter als zuvor beschrieben. Dies hat auch einen bestimmten Grund: Den Einsatz einer speziellen Funktion aufzeigen, sowie auch wann und wieso diese vorkommt. Welche spezielle Funktion? Fragst du dich wahrscheinlich in diesem Moment. Nun, um diese Frage kurz zu beantworten, es ist ``YouTube's favorite special function'' laut dem Mathematiker Michael Penn, die Lambert-\(W\)-Funktion \(W(x)\) welche im Kapitel \ref{buch:section:lambertw} bereits beschrieben wurde.
+In diesem Abschnitt werden algebraische Umformungen ein wenig detaillierter als zuvor beschrieben. Dies hat auch einen bestimmten Grund: Den Einsatz einer speziellen Funktion aufzeigen, sowie auch wann und wieso diese vorkommt. Welche spezielle Funktion? Fragt man sich wahrscheinlich in diesem Moment. Nun, um diese Frage kurz zu beantworten, es ist ``YouTube's favorite special function'' laut dem Mathematiker Michael Penn, die Lambert-\(W\)-Funktion \(W(x)\) welche im Kapitel \ref{buch:section:lambertw} bereits beschrieben wurde.
 
 \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}\biggl(\left(y_0+r_0\right)\frac{x}{x_0^2}+\left(y_0-r_0\right)\frac{1}{x}\biggr).
 	\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
 	=
@@ -348,50 +361,61 @@ 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}
-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. 
+\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:
+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 \(\:1 / (r_0-y_0)\:\) potenzieren:
 \begin{equation}
 	\operatorname{exp}\left(\displaystyle \frac{-4t}{r_0-y_0}+\frac{y_0+r_0}{r_0-y_0}\right)
 	=
 	\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:
+
+\subsubsection{Eine essenzielle Substitution
+	\label{lambertw:subsubsection:SubstChi}}
+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:
+
+\subsubsection{Funktion nach der Zeit dank Lambert-\(W\)
+	\label{lambertw:subsubsection:LambertWundFvonT}}
+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 +426,14 @@ 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}\biggl(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\biggl(\biggl(\frac{x(t)}{x_0}\biggr)^2\biggr)-r_0+3y_0\biggr).
 	\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 +443,4 @@ 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/parzyl/images/Makefile b/buch/papers/parzyl/images/Makefile
new file mode 100644
index 0000000..4bd13ec
--- /dev/null
+++ b/buch/papers/parzyl/images/Makefile
@@ -0,0 +1,16 @@
+#
+# Makefile to build 3d images
+#
+# (c) 2022 Prof Dr Andreas Müller
+#
+
+all:	halfplane.pdf
+
+halfplane.pdf:	halfplane.tex halfplane.jpg
+	pdflatex halfplane.tex
+halfplane.png:	halfplane.pov
+	povray +A0.1 -W1920 -H1080 -Ohalfplane.png halfplane.pov
+halfplane.jpg: halfplane.png Makefile
+	convert -extract 1280x1080+340+0 halfplane.png \
+		-density 300 -units PixelsPerInch halfplane.jpg
+
diff --git a/buch/papers/parzyl/images/common.inc b/buch/papers/parzyl/images/common.inc
new file mode 100644
index 0000000..28aed2b
--- /dev/null
+++ b/buch/papers/parzyl/images/common.inc
@@ -0,0 +1,64 @@
+//
+// common.inc -- some common useful tools for drawing 3d images
+//
+// (c) 2018 Prof Dr Andreas Müller, Hochschule Rapperswil
+//
+
+//
+// draw a right angle quarter circle at point <o> with legs <v1> and <v2> and 
+// color <c>
+//
+#declare rechterwinkelradius = 0.5;
+#declare rechterwinkelthickness = 0.01;
+#macro rechterwinkel(o, v1, v2, c)
+intersection {
+	sphere { o, rechterwinkelradius }
+	#declare rnormale = vnormalize(vcross(v1, v2));
+	plane { rnormale, vdot(o, rnormale) + rechterwinkelthickness * rechterwinkelradius / 0.5 }
+	plane { -rnormale, -vdot(o, rnormale) + rechterwinkelthickness * rechterwinkelradius / 0.5 }
+	plane { -v1, -vdot(o, v1) }
+	plane { -v2, -vdot(o, v2) }
+	pigment {
+		color c
+	}
+}
+sphere { o + 0.45 * (vnormalize(v1) +vnormalize(v2)) * rechterwinkelradius,
+	0.05 * rechterwinkelradius / 0.5
+	pigment {
+		color c
+	}
+}
+#end
+
+//
+// draw an arrow from <from> to <to> with thickness <arrowthickness> with
+// color <c>
+//
+#macro arrow(from, to, arrowthickness, c)
+        #declare arrowdirection = vnormalize(to - from);
+        #declare arrowlength = vlength(to - from);
+        union { 
+                sphere {
+                        from, 1.1 * arrowthickness
+                }
+                cylinder {
+                        from,
+                        from + (arrowlength - 5 * arrowthickness) * arrowdirection,                     
+                        arrowthickness
+                }
+                cone {  
+                        from + (arrowlength - 5 * arrowthickness) * arrowdirection,                     
+                        2 * arrowthickness,
+                        to,
+                        0
+                }
+                pigment {
+                        color c
+                }
+                finish {
+                        specular 0.9
+                        metallic
+                }
+        }
+#end
+
diff --git a/buch/papers/parzyl/images/halfplane.jpg b/buch/papers/parzyl/images/halfplane.jpg
new file mode 100644
index 0000000..8cb5ae3
--- /dev/null
+++ b/buch/papers/parzyl/images/halfplane.jpg
diff --git a/buch/papers/parzyl/images/halfplane.pdf b/buch/papers/parzyl/images/halfplane.pdf
new file mode 100644
index 0000000..7275810
--- /dev/null
+++ b/buch/papers/parzyl/images/halfplane.pdf
diff --git a/buch/papers/parzyl/images/halfplane.png b/buch/papers/parzyl/images/halfplane.png
new file mode 100644
index 0000000..5beefa0
--- /dev/null
+++ b/buch/papers/parzyl/images/halfplane.png
diff --git a/buch/papers/parzyl/images/halfplane.pov b/buch/papers/parzyl/images/halfplane.pov
new file mode 100644
index 0000000..419bb67
--- /dev/null
+++ b/buch/papers/parzyl/images/halfplane.pov
@@ -0,0 +1,201 @@
+//
+// 3dimage.pov
+//
+// (c) 2022 Prof Dr Andreas Müller
+//
+#version 3.7;
+#include "colors.inc"
+#include "skies.inc"
+#include "common.inc"
+
+global_settings {
+        assumed_gamma 1
+}
+
+#declare imagescale = 0.63;
+#declare ar = 0.02;
+
+#declare Cameracenter = <5,3,-4>;
+#declare Worldpoint = <0,-0.80, 0>;
+#declare Lightsource = < 7,10,-3>;
+#declare Lightdirection = vnormalize(Lightsource - Worldpoint);
+#declare Lightaxis1 = vnormalize(vcross(Lightdirection, <0,1,0>));
+#declare Lightaxis2 = vnormalize(vcross(Lightaxis1, Lightdirection));
+
+camera {
+	location Cameracenter
+	look_at Worldpoint
+	right 16/9 * x * imagescale
+	up y * imagescale
+}
+
+light_source {
+	Lightsource color White
+	area_light Lightaxis1 Lightaxis2, 10, 10
+	adaptive 1
+	jitter
+}
+
+sky_sphere {
+	pigment {
+		color White
+	}
+}
+
+arrow( <-2.1, 0,   0 >, < 2.2, 0,   0 >, ar, White)
+arrow( < 0,  -1.1, 0 >, < 0, 1.3, 0 >, ar, White)
+arrow( < 0,   0,  -2 >, < 0, 0,   2.2 >, ar, White)
+
+#declare planecolor = rgb<0.2,0.6,1.0>;
+#declare r = 0.01;
+
+#macro planebox()
+	box { <-2.1,-1.1,-2.1>, <0,1.1,2.1> }
+#end
+
+intersection {
+	plane { <0, 0,  1>, 0.001 }
+	plane { <0, 0, -1>, 0.001 }
+	planebox()
+	pigment {
+		color planecolor transmit 0.3
+	}
+	finish {
+		metallic
+		specular 0.95
+	}
+}
+
+#declare Xstep = 0.2;
+
+intersection {
+	union {
+		#declare X = 0;
+		#while (X > -2.5)
+			cylinder { <X,-3,0>, <X,+3,0>, r }
+			#declare X = X - Xstep;
+		#end
+
+		#declare Y = Xstep;
+		#while (Y < 2.5)
+			cylinder { <-3,  Y, 0>, <0,  Y, 0>, r }
+			cylinder { <-3, -Y, 0>, <0, -Y, 0>, r }
+			#declare Y = Y + Xstep;
+		#end
+	}
+	planebox()
+	pigment {
+		color planecolor
+	}
+	finish {
+		metallic
+		specular 0.95
+	}
+}
+
+#declare parammin = -4;
+#declare parammax = 4;
+#declare paramsteps = 100;
+#declare paramstep = (parammax - parammin) / paramsteps;
+
+#macro punkt(sigma, tau, Z)
+	<
+		0.5 * (tau*tau - sigma*sigma)
+		Z,
+		sigma * tau,
+	>
+#end
+
+#macro sigmasurface(sigma, farbe)
+	#declare taumin1 = 2/sigma;
+	#declare taumin2 = sqrt(4+sigma*sigma);
+	#if (taumin1 > taumin2)
+		#declare taumin = -taumin2;
+	#else
+		#declare taumin = -taumin1;
+	#end
+	
+	mesh {
+		#declare tau = taumin;
+		#declare taumax = -taumin;
+		#declare taustep = (taumax - taumin) / paramsteps;
+		#while (tau < taumax - taustep/2)
+			triangle {
+				punkt(sigma, tau,           -1),
+				punkt(sigma, tau,            0),
+				punkt(sigma, tau + taustep, -1)
+			}
+			triangle {
+				punkt(sigma, tau + taustep, -1),
+				punkt(sigma, tau + taustep,  0),
+				punkt(sigma, tau,            0)
+			}
+			#declare tau = tau + taustep;
+		#end
+		pigment {
+			color farbe
+		}
+		finish {
+			specular 0.9
+			metallic
+		}
+	}
+
+	union {
+		#declare tau = taumin;
+		#declare taumax = -taumin;
+		#declare taustep = (taumax - taumin) / paramsteps;
+		#while (tau < taumax - taustep/2)
+			sphere { punkt(sigma, tau, 0), r }
+			cylinder {
+				punkt(sigma, tau,           0),
+				punkt(sigma, tau + taustep, 0),
+				r
+			}
+			#declare tau = tau + taustep;
+		#end
+		sphere { punkt(sigma, tau, 0), r }
+		pigment {
+			color farbe
+		}
+		finish {
+			specular 0.9
+			metallic
+		}
+		
+	}
+#end
+
+#declare greensurfacecolor = rgb<0.6,1.0,0.6>;
+#declare redsurfacecolor = rgb<1.0,0.6,0.6>;
+
+sigmasurface(0.25, greensurfacecolor)
+sigmasurface(0.5,  greensurfacecolor)
+sigmasurface(0.75, greensurfacecolor)
+sigmasurface(1,    greensurfacecolor)
+sigmasurface(1.25, greensurfacecolor)
+sigmasurface(1.5,  greensurfacecolor)
+sigmasurface(1.75, greensurfacecolor)
+sigmasurface(2,    greensurfacecolor)
+
+union {
+	sigmasurface(0.25, redsurfacecolor)
+	sigmasurface(0.5,  redsurfacecolor)
+	sigmasurface(0.75, redsurfacecolor)
+	sigmasurface(1.00, redsurfacecolor)
+	sigmasurface(1.25, redsurfacecolor)
+	sigmasurface(1.5,  redsurfacecolor)
+	sigmasurface(1.75, redsurfacecolor)
+	sigmasurface(2,    redsurfacecolor)
+	rotate <0, 180, 0>
+}
+
+box { <-2,-1,-2>, <2,-0.99,2> 
+	pigment {
+		color rgb<1.0,0.8,0.6> transmit 0.8
+	}
+	finish {
+		specular 0.9
+		metallic
+	}
+}
diff --git a/buch/papers/parzyl/images/halfplane.tex b/buch/papers/parzyl/images/halfplane.tex
new file mode 100644
index 0000000..e470057
--- /dev/null
+++ b/buch/papers/parzyl/images/halfplane.tex
@@ -0,0 +1,41 @@
+%
+% halfplane.tex
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\documentclass[tikz]{standalone}
+\usepackage{times}
+\usepackage{amsmath}
+\usepackage{txfonts}
+\usepackage[utf8]{inputenc}
+\usepackage{graphics}
+\usetikzlibrary{arrows,intersections,math}
+\usepackage{ifthen}
+\begin{document}
+
+\newboolean{showgrid}
+\setboolean{showgrid}{false}
+\def\breite{5}
+\def\hoehe{4}
+
+\begin{tikzpicture}[>=latex,thick]
+
+% Povray Bild
+\node at (0,0) {\includegraphics[width=10cm]{halfplane.jpg}};
+
+% Gitter
+\ifthenelse{\boolean{showgrid}}{
+\draw[step=0.1,line width=0.1pt] (-\breite,-\hoehe) grid (\breite, \hoehe);
+\draw[step=0.5,line width=0.4pt] (-\breite,-\hoehe) grid (\breite, \hoehe);
+\draw                            (-\breite,-\hoehe) grid (\breite, \hoehe);
+\fill (0,0) circle[radius=0.05];
+}{}
+
+\node at (0,3.7) {$z$};
+\node at (3.3,-0.3) {$x$};
+\node at (2.7,2.5) {$y$};
+
+\end{tikzpicture}
+
+\end{document}
+
diff --git a/buch/papers/sturmliouville/.gitignore b/buch/papers/sturmliouville/.gitignore
new file mode 100644
index 0000000..f08278d
--- /dev/null
+++ b/buch/papers/sturmliouville/.gitignore
@@ -0,0 +1 @@
+*.pdf
\ No newline at end of file
diff --git a/buch/papers/sturmliouville/Makefile b/buch/papers/sturmliouville/Makefile
index da902e7..8d3e0af 100644
--- a/buch/papers/sturmliouville/Makefile
+++ b/buch/papers/sturmliouville/Makefile
@@ -1,9 +1,37 @@
 #
-# Makefile -- make file for the paper sturmliouville
+# Makefile -- make file for the paper fm
 #
 # (c) 2020 Prof Dr Andreas Mueller
 #
 
-images:
-	@echo "no images to be created in sturmliouville"
+SOURCES := \
+	einleitung.tex\
+	eigenschaften.tex \
+	beispiele.tex \
+	main.tex 
 
+#TIKZFIGURES := \
+	tikz/atoms-grid-still.tex \
+
+#FIGURES := $(patsubst tikz/%.tex, figures/%.pdf, $(TIKZFIGURES))
+
+#.PHONY: images
+#images: $(FIGURES)
+
+#figures/%.pdf: tikz/%.tex
+#	mkdir -p figures
+#	pdflatex --output-directory=figures $<
+
+.PHONY: standalone
+standalone: standalone.tex $(SOURCES)  #$(FIGURES)
+	mkdir -p standalone
+	cd ../..; \
+		pdflatex \
+			--halt-on-error \
+			--shell-escape \
+			--output-directory=papers/sturmliouville/standalone \
+			--extra-mem-top=10000000 \
+			papers/sturmliouville/standalone.tex;
+	cd standalone; \
+		bibtex standalone; \
+		makeindex standalone;
\ No newline at end of file
diff --git a/buch/papers/sturmliouville/Makefile.inc b/buch/papers/sturmliouville/Makefile.inc
index e2039ce..7ffdad2 100644
--- a/buch/papers/sturmliouville/Makefile.inc
+++ b/buch/papers/sturmliouville/Makefile.inc
@@ -3,12 +3,12 @@
 #
 # (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
 #
-dependencies-sturmliouville =						\
+dependencies-sturmliouville =							\
 	papers/sturmliouville/packages.tex					\
-	papers/sturmliouville/main.tex					\
+	papers/sturmliouville/main.tex						\
 	papers/sturmliouville/references.bib				\
-	papers/sturmliouville/teil0.tex					\
-	papers/sturmliouville/teil1.tex					\
-	papers/sturmliouville/teil2.tex					\
-	papers/sturmliouville/teil3.tex
-
+	papers/sturmliouville/einleitung.tex				\
+	papers/sturmliouville/eigenschaften.tex				\
+	papers/sturmliouville/beispiele.tex					\
+	papers/sturmliouville/waermeleitung_beispiel.tex	\
+	papers/sturmliouville/tschebyscheff_beispiel.tex
diff --git a/buch/papers/sturmliouville/beispiele.tex b/buch/papers/sturmliouville/beispiele.tex
new file mode 100644
index 0000000..94082cf
--- /dev/null
+++ b/buch/papers/sturmliouville/beispiele.tex
@@ -0,0 +1,14 @@
+%
+% teil2.tex -- Beispiel-File für teil2 
+%
+% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
+%
+\section{Beispiele 
+\label{sturmliouville:section:examples}}
+\rhead{Beispiele}
+
+% Fourier: Erik work
+\input{papers/sturmliouville/waermeleitung_beispiel.tex}
+
+% Tschebyscheff
+\input{papers/sturmliouville/tschebyscheff_beispiel.tex}
\ No newline at end of file
diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex
new file mode 100644
index 0000000..bef8a39
--- /dev/null
+++ b/buch/papers/sturmliouville/eigenschaften.tex
@@ -0,0 +1,83 @@
+%
+% eigenschaften.tex -- Eigenschaften der Lösungen
+% Author: Erik Löffler
+%
+% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
+%
+\section{Eigenschaften von Lösungen
+\label{sturmliouville:section:solution-properties}}
+\rhead{Eigenschaften von Lösungen}
+
+Im weiteren werden nun die Eigenschaften der Lösungen eines
+Sturm-Liouville-Problems diskutiert und aufgezeigt, wie diese Eigenschaften
+zustande kommen.
+
+Dazu wird der Operator $L_0$ welcher bereits in 
+Kapitel~\ref{buch:integrale:subsection:sturm-liouville-problem} betrachtet
+wurde, noch etwas genauer angeschaut.
+Es wird also im Folgenden
+\[
+    L_0
+    =
+    -\frac{d}{dx}p(x)\frac{d}{dx}
+\]
+zusammen mit den Randbedingungen
+\[
+    \begin{aligned}
+        k_a y(a) + h_a p(a) y'(a) &= 0 \\
+        k_b y(b) + h_b p(b) y'(b) &= 0
+    \end{aligned}
+\]
+verwendet.
+Wie im Kapitel~\ref{buch:integrale:subsection:sturm-liouville-problem} bereits 
+gezeigt, resultieren die Randbedingungen aus der Anforderung den Operator $L_0$
+selbsadjungiert zu machen.
+Es wurde allerdings noch nicht darauf eingegangen, welche Eigenschaften dies
+für die Lösungen des Sturm-Liouville-Problems zur Folge hat.
+
+\subsubsection{Exkurs zum Spektralsatz}
+
+Um zu verstehen welche Eigenschaften der selbstadjungierte Operator $L_0$ in 
+den Lösungen hervorbringt, wird der Spektralsatz benötigt.
+
+Dieser wird in der linearen Algebra oft verwendet um zu zeigen, dass eine Matrix
+diagonalisierbar ist, beziehungsweise dass eine Orthonormalbasis existiert.
+
+Im Fall einer gegebenen $n\times n$-Matrix $A$ mit reellen Einträgen wird dazu 
+zunächst gezeigt, dass $A$ selbstadjungiert ist, also dass
+\[
+    \langle Av, w \rangle
+    =
+    \langle v, Aw \rangle
+\]
+für $ v, w \in \mathbb{R}^n$ gilt.
+Ist dies der Fall, kann die Aussage des Spektralsatzes
+\cite{sturmliouville:spektralsatz-wiki} verwended werden.
+Daraus folgt dann, dass eine Orthonormalbasis aus Eigenvektoren existiert,
+wenn $A$ nur Eigenwerte aus $\mathbb{R}$ besitzt.
+
+Dies ist allerdings nicht die Einzige Version des Spektralsatzes.
+Unter anderen gibt es den Spektralsatz für kompakte Operatoren
+\cite{sturmliouville:spektralsatz-wiki}, welcher für das
+Sturm-Liouville-Problem von Bedeutung ist.
+Welche Voraussetzungen erfüllt sein müssen, um diese Version des
+Satzes verwenden zu können, wird hier aber nicht diskutiert und kann bei den
+Beispielen in diesem Kapitel als gegeben betrachtet werden.
+Grundsätzlich ist die Aussage in dieser Version dieselbe, wie bei den Matrizen,
+also dass für ein Operator eine Orthonormalbasis aus Eigenvektoren existiert,
+falls er selbstadjungiert ist.
+
+\subsubsection{Anwendung des Spektralsatzes auf $L_0$}
+
+Der Spektralsatz besagt also, dass, weil $L_0$ selbstadjungiert ist, eine
+Orthonormalbasis aus Eigenvektoren existiert.
+Genauer bedeutet dies, dass alle Eigenvektoren, beziehungsweise alle Lösungen
+des Sturm-Liouville-Problems orthogonal zueinander sind bezüglich des
+Skalarprodukts, in dem $L_0$ selbstadjungiert ist.
+
+Erfüllt also eine Differenzialgleichung die in
+Abschnitt~\ref{sturmliouville:section:teil0} präsentierten Eigenschaften und 
+erfüllen die Randbedingungen der Differentialgleichung die Randbedingungen
+des Sturm-Liouville-Problems, kann bereits geschlossen werden, dass die
+Lösungsfunktion des Problems eine Linearkombination aus orthogonalen
+Basisfunktionen ist.
\ No newline at end of file
diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex
new file mode 100644
index 0000000..d497622
--- /dev/null
+++ b/buch/papers/sturmliouville/einleitung.tex
@@ -0,0 +1,136 @@
+%
+% einleitung.tex -- Beispiel-File für die Einleitung
+%
+% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
+%
+\section{Was ist das Sturm-Liouville-Problem\label{sturmliouville:section:teil0}}
+\rhead{Einleitung}
+Das Sturm-Liouville-Problem wurde benannt nach dem schweizerisch-französischen Mathematiker und Physiker Jacques Charles Fran\c{c}ois Sturm und dem französischen Mathematiker Joseph Liouville.
+Gemeinsam haben sie in der mathematischen Physik die Sturm-Liouville-Theorie entwickelt und gilt für die Lösung von gewöhnlichen Differentialgleichungen, jedoch verwendet man die Theorie öfters bei der Lösung von partiellen Differentialgleichungen.
+Normalerweise betrachtet man für das Strum-Liouville-Problem eine gewöhnliche Differentialgleichung 2. Ordnung, und wenn es sich um eine partielle Differentialgleichung handelt, kann man sie in mehrere gewöhnliche Differentialgleichungen umwandeln. Wie z. B. den Separationsansatz, die partielle Differentialgleichung mit mehreren Variablen.
+
+\begin{definition}
+	\index{Sturm-Liouville-Gleichung}%
+Wenn die lineare homogene Differentialgleichung
+\begin{equation}
+	\frac{d^2y}{dx^2} + a(x)\frac{dy}{dx} + b(x)y = 0
+\end{equation}
+als
+\begin{equation}
+	\label{eq:sturm-liouville-equation}
+	\frac{d}{dx}\lbrack p(x) \frac{dy}{dx} \rbrack + \lbrack q(x) + \lambda w(x) \rbrack y = 0 
+\end{equation}
+geschrieben werden kann, dann wird diese Gleichung als Sturm-Liouville-Gleichung bezeichnet.
+\end{definition}
+Alle homogene 2. Ordnung lineare gewöhnliche Differentialgleichungen können in die Form der Gleichung \ref{eq:sturm-liouville-equation} umgeformt werden.
+
+\subsection{Randbedingungen\label{sub:was-ist-das-slp-randbedingungen}}
+Wenn von der Funktion $y(x)$ die Werte $x$ des jeweiligen Randes des Definitionsbereiches anzunehmen sind, also
+\begin{equation}
+	y(a) = y(b) = 0,
+\end{equation}
+so spricht man von einer Dirichlet-Randbedingung\footnote{Die Dirichlet-Randbedingung oder auch Randbedingung des ersten Typs genannt ist nach dem deutschen Mathematiker Peter Gstav Lejeune Dirichlet benannt. Sie findet Anwendung auf gewöhnliche oder patielle Differentialgleichungen und gibt mit der Bedingung die Werte an, die für die abgeleitete Lösung innerhalb der Domänengrenze gelten.}, und von einer Neumann-Randbedingung\footnote{Die Neumann-Randbedingung oder auch Randbedingung des zweiten Typs genannt, ist nach dem deutschen Mathematiker Carl Neumann benannt. Sie legt die Werte fest, die eine Lösung entlang der Domänengrenze annehmen muss, wenn eine gewöhnliche oder partielle Differentialgleichung gestellt wird.} spricht man, wenn
+\begin{equation}
+	y'(a) = y'(b) = 0
+\end{equation}
+ergibt.
+
+Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung mit den homogenen Randbedingungen des dritten Typs\footnote{Die Randbedingung des dritten Typs, oder Robin-Randbedingungen (benannt nach dem französischen mathematischen Analytiker und angewandten Mathematiker Victor Gustave Robin), wird genannt, wenn sie einer gewöhnlichen oder partiellen Differentialgleichung auferlegt wird, so sind die Spezifikationen einer Linearkombination der Werte einer Funktion sowie die Werte ihrer Ableitung am Rande des Bereichs}
+\begin{equation}
+\begin{aligned}
+	\label{eq:randbedingungen}
+	k_a y(a) + h_a p(a) y'(a) &= 0 \\
+	k_b y(b) + h_b p(b) y'(b) &= 0
+\end{aligned}
+\end{equation}
+kombiniert, dann bekommt man das klassische Sturm-Liouville-Problem.
+
+\subsection{Eigenwertproblem}
+Die Gleichungen \ref{eq:sturm-liouville-equation} hat die Form eines Eigenwertproblems
+Wenn bei der Sturm-Liouville-Gleichung \ref{eq:sturm-liouville-equation} alles  konstant bleibt, aber der Wert von $\lambda$ sich ändert, erhält man eine andere Eigenfunktion, weil man eine andere gewöhnliche Differentialgleichung löst;
+der Parameter $\lambda$ wird als Eigenwert bezeichnet.
+Es ist genau das gleiche Prinzip wie bei den Matrizen, andere Eigenwerte ergeben andere Eigenvektoren.
+Es besteht eine Korrespondenz zwischen den Eigenwerten und den Eigenvektoren.
+Das gleiche gilt auch beim Sturm-Liouville-Problem, und zwar
+\begin{equation}
+	\lambda \overset{Korrespondenz}\leftrightarrow y.
+\end{equation}
+
+Die Theorie besagt, wenn $y_m$, $y_n$ Eigenfuktionen des Sturm-Liouville-Problems sind, die verschiedene Eigenwerte $\lambda_m$, $\lambda_n$ ($\lambda_m \neq \lambda_n$) entsprechen, so sind $y_m$, $y_n$ orthogonal zu y -
+dies gilt für das Intervall (a,b).
+Somit ergibt die Gleichung
+\begin{equation}
+	\label{eq:skalar-sturm-liouville}
+	\int_{a}^{b} w(x)y_m y_n = 0.
+\end{equation}
+
+\subsection{Koeffizientenfunktionen}
+Die Funktionen $p(x)$, $q(x)$ und $w(x)$ werden als Koeffizientenfunktionen mit ihren freien Variablen $x$ bezeichnet.
+Die Funktion $w(x)$ (manchmal auch $r(x)$ genannt) wird als Gewichtsfunktion oder Dichtefunktion bezeichnet.
+Es gibt zwei verschiedene Sturm-Liouville-Probleme: das reguläre Sturm-Liouville-Problem und das singuläre Sturm-Liouville-Problem. 
+Die Funktionen für das reguläre und das singuläre Sturm-Liouville-Problem sind nicht dieselben.
+
+%
+%Kapitel mit "Das reguläre Sturm-Liouville-Problem"
+%
+
+\subsection{Das reguläre Sturm-Liouville-Problem\label{sub:reguläre_sturm_liouville_problem}}
+Damit es sich um ein reguläres Sturm-Liouville-Problem handelt, müssen einige Bedingungen beachtet werden.
+\begin{definition}
+	\label{def:reguläres_sturm-liouville-problem}
+	\index{regläres Sturm-Liouville-Problem}
+	Die Bedingungen für ein reguläres Sturm-Liouville-Problem sind:
+	\begin{itemize}
+		\item Die Funktionen $p(x), p'(x), q(x)$ und $w(x)$ müssen stetig und reell sein.
+		\item sowie müssen in einem endlichen Intervall $[a,b]$ integrierbar sein.
+		\item $p(x)$ und $w(x)$ sind $>0$.
+		\item Es gelten die Randbedingungen \ref{eq:randbedingungen}, wobei $|k_i|^2 + |h_i|^2\ne 0$ mit $i=a,b$.
+	\end{itemize}
+\end{definition}
+Bei einem regulären Sturm-Liouville-Problem geht es darum, wichtige Eigenschaften der Eigenfunktionen beschreiben zu können, ohne sie genau zu kennen.
+
+
+%
+%Kapitel mit "Das singuläre Sturm-Liouville-Problem"
+%
+
+
+\subsection{Das singuläre Sturm-Liouville-Problem\label{sub:singuläre_sturm_liouville_problem}}
+Von einem singulären Sturm-Liouville-Problem spricht man, wenn die Bedingungen des regulärem Problem nicht erfüllt sind.
+\begin{definition}
+	\label{def:singulär_sturm-liouville-problem}
+	\index{singuläres Sturm-Liouville-Problem}
+Es handelt sich um ein singuläres Sturm-Liouville-Problem, wenn:
+	\begin{itemize}
+		\item wenn sein Definitionsbereich auf dem Intervall $[ \ a,b] \ $ unbeschränkt ist oder
+		\item wenn die Koeffizienten an den Randpunkten Singularitäten haben.
+	\end{itemize}
+\end{definition}
+Allerdings kann nur eine der Bedingungen nicht erfüllt sein, so dass es sich bereits um ein singuläres Sturm-Liouville-Problem handelt.
+
+\begin{beispiel}
+	Das Randwertproblem
+	\begin{equation}
+		\begin{aligned}
+		x^2y'' + xy' + (\lambda^2x^2 - m^2)y &= 0, 0<x<a,\\
+		y(a) &= 0
+		\end{aligned}
+	\end{equation}
+	ist kein reguläres Sturm-Liouville-Problem.
+	Wenn man die Gleichung in die Sturm-Liouville Form umformen, dann ergeben die Koeffizientenfunktionen $p(x) = w(x) = x$ und $q(x) = -m^2/x$.
+	Schaut man jetzt die Bedingungen im Kapitel \ref{sub:reguläre_sturm_liouville_problem} an und vergleicht diese unseren Koeffizientenfunktionen, so erkennt man einige Probleme:
+	\begin{itemize}
+		\item $p(x)$ und $w(x)$ sind nicht positiv, wenn $x = 0$ ist.
+		\item $q(x)$ ist nicht kontinuierlich, wenn $x = 0$ ist.
+		\item Die Randbedingung bei $x = 0$ fehlt.
+	\end{itemize}
+\end{beispiel}
+
+Verwendet man das reguläre Sturm-Liouville-Problem, obwohl eine oder beide Bedingungen nicht erfüllt sind, dann ist es schwierig zu sagen, ob die Lösung eindeutige Ergebnisse hat.
+Es ist schwierig, Kriterien anzuwenden, da die Formulierungen z. B. in der Lösungsfunktion liegen.
+Ähnlich wie bei der Fourier-Reihe gegenüber der Fourier-Transformation gibt es immer noch eine zugehörige Eigenfunktionsentwicklung, und zwar die Integraltransformation sowie gibt es weiterhin verallgemeinerte Eigenfunktionen.
+
+ 
+
+
+
diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex
index a7d2857..4b5b8af 100644
--- a/buch/papers/sturmliouville/main.tex
+++ b/buch/papers/sturmliouville/main.tex
@@ -1,36 +1,20 @@
+% !TeX root = ../../buch.tex
 %
 % main.tex -- Paper zum Thema <sturmliouville>
 %
 % (c) 2020 Hochschule Rapperswil
 %
-\chapter{Thema\label{chapter:sturmliouville}}
-\lhead{Thema}
+\chapter{Sturm-Liouville-Problem\label{chapter:sturmliouville}}
+\lhead{Sturm-Liouville-Problem}
 \begin{refsection}
-\chapterauthor{Hans Muster}
+\chapterauthor{Réda Haddouche und Erik Löffler}
 
-Ein paar Hinweise für die korrekte Formatierung des Textes
-\begin{itemize}
-\item
-Absätze werden gebildet, indem man eine Leerzeile einfügt.
-Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet.
-\item
-Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende
-Optionen werden gelöscht. 
-Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen.
-\item
-Beginnen Sie jeden Satz auf einer neuen Zeile. 
-Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen
-in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt 
-anzuwenden.
-\item 
-Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren
-Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern.
-\end{itemize}
-
-\input{papers/sturmliouville/teil0.tex}
-\input{papers/sturmliouville/teil1.tex}
-\input{papers/sturmliouville/teil2.tex}
-\input{papers/sturmliouville/teil3.tex}
+\input{papers/sturmliouville/einleitung.tex}
+%einleitung "was ist das sturm-liouville-problem"
+\input{papers/sturmliouville/eigenschaften.tex}
+%Eigenschaften von Lösungen eines solchen Problems
+\input{papers/sturmliouville/beispiele.tex}
+%Beispiele sind: Wärmeleitung in einem Stab, Tschebyscheff-Polynome
 
 \printbibliography[heading=subbibliography]
 \end{refsection}
diff --git a/buch/papers/sturmliouville/references.bib b/buch/papers/sturmliouville/references.bib
index f66a74d..0c4724b 100644
--- a/buch/papers/sturmliouville/references.bib
+++ b/buch/papers/sturmliouville/references.bib
@@ -4,6 +4,19 @@
 % (c) 2020 Autor, Hochschule Rapperswil
 %
 
+@online{sturmliouville:spektralsatz-wiki,
+	title = {Spektralsatz},
+	url = {https://de.wikipedia.org/wiki/Spektralsatz},
+	date = {2020-08-15},
+	year = {2020},
+	month = {8},
+	day = {15}
+}
+
+%
+% examples (not referenced in book)
+%
+
 @online{sturmliouville:bibtex,
 	title = {BibTeX},
 	url = {https://de.wikipedia.org/wiki/BibTeX},
diff --git a/buch/papers/sturmliouville/teil0.tex b/buch/papers/sturmliouville/teil0.tex
deleted file mode 100644
index ffcb8f3..0000000
--- a/buch/papers/sturmliouville/teil0.tex
+++ /dev/null
@@ -1,22 +0,0 @@
-%
-% einleitung.tex -- Beispiel-File für die Einleitung
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 0\label{sturmliouville:section:teil0}}
-\rhead{Teil 0}
-Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam
-nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam
-erat, sed diam voluptua \cite{sturmliouville:bibtex}.
-At vero eos et accusam et justo duo dolores et ea rebum.
-Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum
-dolor sit amet.
-
-Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam
-nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam
-erat, sed diam voluptua.
-At vero eos et accusam et justo duo dolores et ea rebum.  Stet clita
-kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit
-amet.
-
-
diff --git a/buch/papers/sturmliouville/teil1.tex b/buch/papers/sturmliouville/teil1.tex
deleted file mode 100644
index c23c1d6..0000000
--- a/buch/papers/sturmliouville/teil1.tex
+++ /dev/null
@@ -1,55 +0,0 @@
-%
-% teil1.tex -- Beispiel-File für das Paper
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 1
-\label{sturmliouville:section:teil1}}
-\rhead{Problemstellung}
-Sed ut perspiciatis unde omnis iste natus error sit voluptatem
-accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
-quae ab illo inventore veritatis et quasi architecto beatae vitae
-dicta sunt explicabo.
-Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit
-aut fugit, sed quia consequuntur magni dolores eos qui ratione
-voluptatem sequi nesciunt
-\begin{equation}
-\int_a^b x^2\, dx
-=
-\left[ \frac13 x^3 \right]_a^b
-=
-\frac{b^3-a^3}3.
-\label{sturmliouville:equation1}
-\end{equation}
-Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet,
-consectetur, adipisci velit, sed quia non numquam eius modi tempora
-incidunt ut labore et dolore magnam aliquam quaerat voluptatem.
-
-Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis
-suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur?
-Quis autem vel eum iure reprehenderit qui in ea voluptate velit
-esse quam nihil molestiae consequatur, vel illum qui dolorem eum
-fugiat quo voluptas nulla pariatur?
-
-\subsection{De finibus bonorum et malorum
-\label{sturmliouville:subsection:finibus}}
-At vero eos et accusamus et iusto odio dignissimos ducimus qui
-blanditiis praesentium voluptatum deleniti atque corrupti quos
-dolores et quas molestias excepturi sint occaecati cupiditate non
-provident, similique sunt in culpa qui officia deserunt mollitia
-animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}.
-
-Et harum quidem rerum facilis est et expedita distinctio
-\ref{sturmliouville:section:loesung}.
-Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil
-impedit quo minus id quod maxime placeat facere possimus, omnis
-voluptas assumenda est, omnis dolor repellendus
-\ref{sturmliouville:section:folgerung}.
-Temporibus autem quibusdam et aut officiis debitis aut rerum
-necessitatibus saepe eveniet ut et voluptates repudiandae sint et
-molestiae non recusandae.
-Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis
-voluptatibus maiores alias consequatur aut perferendis doloribus
-asperiores repellat.
-
-
diff --git a/buch/papers/sturmliouville/teil2.tex b/buch/papers/sturmliouville/teil2.tex
deleted file mode 100644
index 7fc3d2c..0000000
--- a/buch/papers/sturmliouville/teil2.tex
+++ /dev/null
@@ -1,40 +0,0 @@
-%
-% teil2.tex -- Beispiel-File für teil2 
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 2 
-\label{sturmliouville:section:teil2}}
-\rhead{Teil 2}
-Sed ut perspiciatis unde omnis iste natus error sit voluptatem
-accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
-quae ab illo inventore veritatis et quasi architecto beatae vitae
-dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit
-aspernatur aut odit aut fugit, sed quia consequuntur magni dolores
-eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam
-est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci
-velit, sed quia non numquam eius modi tempora incidunt ut labore
-et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima
-veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam,
-nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure
-reprehenderit qui in ea voluptate velit esse quam nihil molestiae
-consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla
-pariatur?
-
-\subsection{De finibus bonorum et malorum
-\label{sturmliouville:subsection:bonorum}}
-At vero eos et accusamus et iusto odio dignissimos ducimus qui
-blanditiis praesentium voluptatum deleniti atque corrupti quos
-dolores et quas molestias excepturi sint occaecati cupiditate non
-provident, similique sunt in culpa qui officia deserunt mollitia
-animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis
-est et expedita distinctio. Nam libero tempore, cum soluta nobis
-est eligendi optio cumque nihil impedit quo minus id quod maxime
-placeat facere possimus, omnis voluptas assumenda est, omnis dolor
-repellendus. Temporibus autem quibusdam et aut officiis debitis aut
-rerum necessitatibus saepe eveniet ut et voluptates repudiandae
-sint et molestiae non recusandae. Itaque earum rerum hic tenetur a
-sapiente delectus, ut aut reiciendis voluptatibus maiores alias
-consequatur aut perferendis doloribus asperiores repellat.
-
-
diff --git a/buch/papers/sturmliouville/teil3.tex b/buch/papers/sturmliouville/teil3.tex
deleted file mode 100644
index 3aa1b40..0000000
--- a/buch/papers/sturmliouville/teil3.tex
+++ /dev/null
@@ -1,40 +0,0 @@
-%
-% teil3.tex -- Beispiel-File für Teil 3
-%
-% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
-%
-\section{Teil 3
-\label{sturmliouville:section:teil3}}
-\rhead{Teil 3}
-Sed ut perspiciatis unde omnis iste natus error sit voluptatem
-accusantium doloremque laudantium, totam rem aperiam, eaque ipsa
-quae ab illo inventore veritatis et quasi architecto beatae vitae
-dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit
-aspernatur aut odit aut fugit, sed quia consequuntur magni dolores
-eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam
-est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci
-velit, sed quia non numquam eius modi tempora incidunt ut labore
-et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima
-veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam,
-nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure
-reprehenderit qui in ea voluptate velit esse quam nihil molestiae
-consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla
-pariatur?
-
-\subsection{De finibus bonorum et malorum
-\label{sturmliouville:subsection:malorum}}
-At vero eos et accusamus et iusto odio dignissimos ducimus qui
-blanditiis praesentium voluptatum deleniti atque corrupti quos
-dolores et quas molestias excepturi sint occaecati cupiditate non
-provident, similique sunt in culpa qui officia deserunt mollitia
-animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis
-est et expedita distinctio. Nam libero tempore, cum soluta nobis
-est eligendi optio cumque nihil impedit quo minus id quod maxime
-placeat facere possimus, omnis voluptas assumenda est, omnis dolor
-repellendus. Temporibus autem quibusdam et aut officiis debitis aut
-rerum necessitatibus saepe eveniet ut et voluptates repudiandae
-sint et molestiae non recusandae. Itaque earum rerum hic tenetur a
-sapiente delectus, ut aut reiciendis voluptatibus maiores alias
-consequatur aut perferendis doloribus asperiores repellat.
-
-
diff --git a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex
new file mode 100644
index 0000000..3817dc0
--- /dev/null
+++ b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex
@@ -0,0 +1,82 @@
+%
+% tschebyscheff_beispiel.tex
+%
+% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
+%
+
+\subsection{Sind Tschebyscheff-Polynome orthogonal zueinander?\label{sub:tschebyscheff-polynome}}
+\subsubsection*{Definition der Koeffizientenfunktion}
+Im Kapitel \ref{sub:beispiele_sturm_liouville_problem} sind die Koeffizientenfunktionen, die man braucht, schon aufgeliste, und zwar mit
+\begin{align*}
+	w(x) &= \frac{1}{\sqrt{1-x^2}} \\
+	p(x) &= \sqrt{1-x^2} \\
+	q(x) &= 0
+\end{align*}.
+Da die Sturm-Liouville-Gleichung
+\begin{equation}
+	\label{eq:sturm-liouville-equation-tscheby}
+	\frac{d}{dx} (\sqrt{1-x^2} \frac{dy}{dx}) + (0 + \lambda \frac{1}{\sqrt{1-x^2}}) y = 0 
+\end{equation}
+nun mit den Koeffizientenfunktionen aufgestellt werden kann, bleibt die Frage, ob es sich um ein reguläres oder singuläres Sturm-Liouville-Problem handelt.
+
+\subsubsection*{regulär oder singulär?}
+Für das reguläre Problem laut der Definition \ref{def:reguläres_sturm-liouville-problem} muss die funktion $p(x) = \sqrt{1-x^2}$, $p'(x) = -2x$, $q(x) = 0$ und $w(x) = \frac{1}{\sqrt{1-x^2}}$ stetig und reell sein --- und sie sind es auch.
+Auf dem Intervall $(-1,1)$ sind die Tschebyscheff-Polynome erster Art mit Hilfe von Hyperbelfunktionen
+\begin{equation}
+	T_n(x) = \cos n (\arccos x)
+\end{equation}.
+Für $x>1$ und $x<-1$ sehen die Polynome wie folgt aus:
+\begin{equation}
+	T_n(x) = \left\{\begin{array}{ll} \cosh (n \arccos x), & x > 1\\
+		(-1)^n \cosh (n \arccos (-x)), & x<-1 \end{array}\right.
+\end{equation},
+jedoch ist die Orthogonalität nur auf dem Intervall $[ -1, 1]$ sichergestellt.
+Die nächste Bedingung beinhaltet, dass die Funktion $p(x)$ und $w(x)>0$ sein müssen.
+Die Funktion
+\begin{equation*}
+	p(x)^{-1} = \frac{1}{\sqrt{1-x^2}}
+\end{equation*}
+ist die gleiche wie $w(x)$ und erfüllt die Bedingung.
+
+\subsubsection*{Randwertproblem}
+Für die Verifizierung der Randbedingungen benötigt man erneut $p(x)$.
+Da sich die Polynome nur auf dem Intervall $[ -1,1 ]$ orthogonal verhalten, sind $a = -1$ und $b = 1$ gesetzt.
+Beim einsetzen in die Randbedingung \ref{eq:randbedingungen}, erhält man
+\begin{equation}
+\begin{aligned}
+	k_a y(-1) + h_a y'(-1) &= 0
+	k_b y(-1) + h_b y'(-1) &= 0.
+\end{aligned} 
+\end{equation}
+Die Funktion $y(x)$ und $y'(x)$ sind in diesem Fall die Tschebyscheff Polynome (siehe \label{sub:definiton_der_tschebyscheff-Polynome}).
+Es gibt zwei Arten von Tschebyscheff Polynome: die erste Art $T_n(x)$ und die zweite Art $U_n(x)$.
+Jedoch beachtet man in diesem Kapitel nur die Tschebyscheff Polynome erster Art (\ref{eq:tschebyscheff-polynome}).
+Die Funktion $y(x)$ wird nun mit der Funktion $T_n(x)$ ersetzt und für die Verifizierung der Randbedingung wählt man $n=2$.
+Somit erhält man
+\begin{equation}
+	\begin{aligned}
+	k_a T_2(-1) + h_a T_{2}'(-1) &= k_a = 0\\
+	k_b T_2(1) + h_b T_{2}'(1) &= k_b = 0.
+\end{aligned}
+\end{equation}
+Ähnlich wie beim Beispiel der Wärmeleitung in einem homogenen Stab kann man, damit die Bedingung $|k_i|^2 + |h_i|^2\ne 0$ erfüllt ist, können beliebige $h_a \ne 0$ und $h_b \ne 0$ gewählt werden.
+Somit ist erneut gezeigt, dass die Randbedingungen der Tschebyscheff-Polynome auf die Sturm-Liouville-Randbedingungen erfüllt und alle daraus resultierenden Lösungen orthogonal sind.
+
+\begin{beispiel}
+	Die Gleichung \ref{eq:skalar-sturm-liouville} mit $y_m = T_1(x)$ und $y_n(x) = T_2(x)$ eingesetzt sowie $a=-1$ und $b = 1$ ergibt
+	\[
+	\int_{-1}^{1} w(x) x (2x^2-1) dx = 0.
+	\]
+\end{beispiel}
+
+ 
+
+
+
+
+
+
+
+
+
+
diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex
new file mode 100644
index 0000000..a72c562
--- /dev/null
+++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex
@@ -0,0 +1,675 @@
+%
+% waermeleitung_beispiel.tex -- Beispiel Wärmeleitung in homogenem Stab.
+% Author: Erik Löffler
+%
+% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
+%
+
+\subsection{Wärmeleitung in einem Homogenen Stab}
+
+In diesem Abschnitt wird das Problem der Wärmeleitung in einem homogenen Stab
+betrachtet und wie das Sturm-Liouville-Problem bei der Beschreibung dieses
+physikalischen Phänomenes auftritt.
+
+Zunächst wird ein eindimensionaler homogener Stab der Länge $l$ und
+Wärmeleitkoeffizient $\kappa$ betrachtet.
+Es ergibt sich für das Wärmeleitungsproblem
+die partielle Differentialgleichung
+\begin{equation}
+    \label{sturmliouville:eq:example-fourier-heat-equation}
+    \frac{\partial u}{\partial t} =
+    \kappa \frac{\partial^{2}u}{{\partial x}^{2}},
+\end{equation}
+wobei der Stab in diesem Fall auf der $X$-Achse im Intervall $[0,l]$ liegt.
+
+Da diese Differentialgleichung das Problem allgemein für einen homogenen
+Stab beschreibt, werden zusätzliche Bedingungen benötigt, um beispielsweise
+die Lösung für einen Stab zu finden, bei dem die Enden auf konstanter 
+Tempreatur gehalten werden.
+
+%
+% Randbedingungen für Stab mit konstanten Endtemperaturen
+%
+\subsubsection{Randbedingungen für Stab mit Enden auf konstanter Temperatur}
+
+Die Enden des Stabes auf konstanter Temperatur zu halten bedeutet, dass die
+Lösungsfunktion $u(t,x)$ bei $x = 0$ und $x = l$ nur die vorgegebene
+Temperatur zurückgeben darf. Diese wird einfachheitshalber als $0$ angenomen.
+Es folgt nun
+\begin{equation}
+    \label{sturmliouville:eq:example-fourier-boundary-condition-ends-constant}
+    u(t,0)
+    =
+    u(t,l)
+    =
+    0
+\end{equation}
+als Randbedingungen.
+
+%
+% Randbedingungen für Stab mit isolierten Enden
+%
+
+\subsubsection{Randbedingungen für Stab mit isolierten Enden}
+
+Bei isolierten Enden des Stabes können beliebige Temperaturen für $x = 0$ und
+$x = l$ auftreten. In diesem Fall ist es nicht erlaubt, dass Wärme vom Stab
+an die Umgebung oder von der Umgebung an den Stab abgegeben wird.
+
+Aus der Physik ist bekannt, dass Wärme immer von der höheren zur tieferen
+Temperatur fliesst. Um Wärmefluss zu unterdrücken, muss also dafür gesorgt
+werden, dass am Rand des Stabes keine Temperaturdifferenz existiert oder 
+dass die partiellen Ableitungen von $u(t,x)$ nach $x$ bei $x = 0$ und $x = l$
+verschwinden.
+Somit folgen
+\begin{equation}
+    \label{sturmliouville:eq:example-fourier-boundary-condition-ends-isolated}
+    \frac{\partial}{\partial x} u(t, 0)
+    =
+    \frac{\partial}{\partial x} u(t, l)
+    =
+    0
+\end{equation}
+als Randbedingungen.
+
+%
+% Lösung der Differenzialgleichung mittels Separation
+%
+
+\subsubsection{Lösung der Differenzialgleichung}
+
+Da die Lösungsfunktion von zwei Variablen abhängig ist, wird als Lösungsansatz
+die Separationsmethode verwendet.
+Dazu wird 
+\[
+    u(t,x)
+    =
+    T(t)X(x)
+\]
+in die partielle 
+Differenzialgleichung~\eqref{sturmliouville:eq:example-fourier-heat-equation}
+eingesetzt.
+Daraus ergibt sich 
+\[
+    T^{\prime}(t)X(x)
+    =
+    \kappa T(t)X^{\prime \prime}(x)
+\]
+als neue Form.
+
+Nun können alle von $t$ abhängigen Ausdrücke auf die linke Seite, sowie alle
+von $x$ abhängigen Ausdrücke auf die rechte Seite gebracht werden und mittels
+der neuen Variablen $\mu$ gekoppelt werden:
+\[
+    \frac{T^{\prime}(t)}{\kappa T(t)}
+    =
+    \frac{X^{\prime \prime}(x)}{X(x)}
+    =
+    \mu
+\]
+Durch die Einführung von $\mu$ kann das Problem nun in zwei separate
+Differenzialgleichungen aufgeteilt werden:
+\begin{equation}
+    \label{sturmliouville:eq:example-fourier-separated-x}
+    X^{\prime \prime}(x) - \mu X(x)
+    =
+    0
+\end{equation}
+\begin{equation}
+    \label{sturmliouville:eq:example-fourier-separated-t}
+    T^{\prime}(t) - \kappa \mu T(t)
+    =
+    0
+\end{equation}
+
+%
+% Überprüfung Orthogonalität der Lösungen
+%
+
+Es ist an dieser Stelle zu bemerken, dass die Gleichung in $x$ in 
+Sturm-Liouville-Form ist.
+Erfüllen die Randbedingungen des Stab-Problems auch die Randbedingungen des
+Sturm-Liouville-Problems, kann bereits die Aussage getroffen werden, dass alle
+Lösungen für die Gleichung in $x$ orthogonal sein werden.
+
+Da die Bedingungen des Stab-Problem nur Anforderungen an $x$ stellen, können
+diese direkt für $X(x)$ übernomen werden. Es gilt also $X(0) = X(l) = 0$.
+Damit die Lösungen von $X$ orthogonal sind, müssen also die Gleichungen
+\begin{equation}
+\begin{aligned}
+	\label{sturmliouville:eq:example-fourier-randbedingungen}
+	k_a X(a) + h_a p(a) X'(a) &= 0 \\
+	k_b X(b) + h_b p(b) X'(b) &= 0
+\end{aligned}
+\end{equation}
+erfüllt sein und es muss ausserdem
+\begin{equation}
+\begin{aligned}
+    \label{sturmliouville:eq:example-fourier-coefficient-constraints}
+    |k_a|^2 + |h_a|^2 &\neq 0\\
+    |k_b|^2 + |h_b|^2 &\neq 0\\
+\end{aligned}
+\end{equation}
+gelten.
+
+Um zu verifizieren, ob die Randbedingungen erfüllt sind, wird zunächst
+$p(x)$
+benötigt.
+Dazu wird die Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x}
+mit der
+Sturm-Liouville-Form~\eqref{eq:sturm-liouville-equation} verglichen, was zu
+$p(x) = 1$ führt.
+
+Werden nun $p(x)$ und die 
+Randbedingungen~\eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-constant}
+in \eqref{sturmliouville:eq:example-fourier-randbedingungen} eigesetzt, erhält
+man
+\[
+\begin{aligned}
+	k_a y(0) + h_a y'(0) &= h_a y'(0) = 0 \\
+	k_b y(l) + h_b y'(l) &= h_b y'(l) = 0.
+\end{aligned}
+\]
+Damit die Gleichungen erfüllt sind, müssen $h_a = 0$ und $h_b = 0$ sein.
+Zusätzlich müssen aber die 
+Bedingungen~\eqref{sturmliouville:eq:example-fourier-coefficient-constraints}
+erfüllt sein und da $y(0) = 0$ und $y(l) = 0$ sind, können belibige $k_a \neq 0$
+und $k_b \neq 0$ gewählt werden.
+
+Somit ist gezeigt, dass die Randbedingungen des Stab-Problems für Enden auf
+konstanter Temperatur auch die Sturm-Liouville-Randbedingungen erfüllen und
+alle daraus reultierenden Lösungen orthogonal sind.
+Analog dazu kann gezeit werden, dass die Randbedingungen für einen Stab mit
+isolierten Enden ebenfalls die Sturm-Liouville-Randbedingungen erfüllen und
+somit auch zu orthogonalen Lösungen führen.
+
+%
+%   Lösung von X(x), Teil mu
+%
+
+\subsubsection{Lösund der Differentialgleichung in $x$}
+Als erstes wird auf die
+Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x} eingegangen.
+Aufgrund der Struktur der Gleichung
+\[
+    X^{\prime \prime}(x) - \mu X(x)
+    =
+    0
+\]
+wird ein trigonometrischer Ansatz gewählt.
+Die Lösungen für $X(x)$ sind also von der Form
+\[
+    X(x)
+    =
+    A \cos \left( \alpha x\right) + B \sin \left( \beta x\right).
+\]
+
+Dieser Ansatz wird nun solange differenziert, bis alle in
+Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x} enthaltenen
+Ableitungen vorhanden sind.
+Man erhält also
+\[
+    X^{\prime}(x)
+    =
+    - \alpha A \sin \left( \alpha x \right) +
+    \beta B \cos \left( \beta x \right)
+\]
+und
+\[
+    X^{\prime \prime}(x)
+    =
+    -\alpha^{2} A \cos \left( \alpha x \right) -
+    \beta^{2} B \sin \left( \beta x \right).
+\]
+
+Eingesetzt in Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x}
+ergibt dies
+\[
+    -\alpha^{2}A\cos(\alpha x) - \beta^{2}B\sin(\beta x) -
+    \mu\left(A\cos(\alpha x) + B\sin(\beta x)\right)
+    =
+    0
+\]
+und durch umformen somit
+\[
+    -\alpha^{2}A\cos(\alpha x) - \beta^{2}B\sin(\beta x)
+    =
+    \mu A\cos(\alpha x) + \mu B\sin(\beta x).
+\]
+
+Mittels Koeffizientenvergleich von
+\[
+\begin{aligned}
+    -\alpha^{2}A\cos(\alpha x)
+    &=
+    \mu A\cos(\alpha x)
+    \\
+    -\beta^{2}B\sin(\beta x)
+    &=
+    \mu B\sin(\beta x)
+\end{aligned}
+\]
+ist schnell ersichtlich, dass $ \mu = -\alpha^{2} = -\beta^{2} $ gelten muss für
+$ A \neq 0 $ oder $ B \neq 0 $.
+Zur Berechnung von $ \mu $ bleiben also noch  $ \alpha $ und $ \beta $ zu
+bestimmen.
+Dazu werden nochmals die
+Randbedingungen~\eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-constant} 
+und \eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-isolated}
+benötigt.
+
+Da die Koeffizienten $A$ und $B$, sowie die Parameter $\alpha$ uns $\beta$ im
+allgemeninen ungleich $0$ sind, müssen die Randbedingungen durch die
+trigonometrischen Funktionen erfüllt werden.
+
+Es werden nun die 
+Randbedingungen~\eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-constant}
+für einen Stab mit Enden auf konstanter Temperatur in die
+Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x} eingesetzt.
+Betrachten wir zunächst die Bedingung für $x = 0$.
+Dies fürht zu
+\[
+    X(0)
+    =
+    A \cos(0 \alpha) + B \sin(0 \beta)
+    =
+    0.
+\]
+Da $\cos(0) \neq 0$ ist, muss in diesem Fall $A = 0$ gelten.
+Für den zweiten Summanden ist wegen $\sin(0) = 0$ die Randbedingung erfüllt.
+
+Wird nun die zweite Randbedingung für $x = l$ mit $A = 0$ eingesetzt, ergibt
+sich
+\[
+    X(l)
+    =
+    0 \cos(\alpha l) + B \sin(\beta l)
+    =
+    B \sin(\beta l)
+    = 0.
+\]
+
+$\beta$ muss also so gewählt werden, dass $\sin(\beta l) = 0$ gilt.
+Es bleibt noch nach $\beta$ aufzulösen:
+\[
+\begin{aligned}
+    \sin(\beta l) &= 0 \\
+    \beta l &= n \pi \qquad n \in \mathbb{N} \\
+    \beta &= \frac{n \pi}{l} \qquad n \in \mathbb{N}
+\end{aligned}
+\]
+
+Es folgt nun wegen $\mu = -\beta^{2}$, dass
+\[
+    \mu_1 = -\beta^{2} = -\frac{n^{2}\pi^{2}}{l^{2}}
+\]
+sein muss.
+Ausserdem ist zu bemerken, dass dies auch gleich $-\alpha^{2}$ ist.
+Da aber $A = 0$ gilt und der Summand mit $\alpha$ verschwindet, ist dies keine
+Verletzung der Randbedingungen.
+
+Durch alanoges Vorgehen kann nun auch das Problem mit isolierten Enden gelöst
+werden.
+Setzt man nun die 
+Randbedingungen~\eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-isolated}
+in $X^{\prime}$ ein, beginnend für $x = 0$. Es ergibt sich
+\[
+    X^{\prime}(0)
+    =
+    -\alpha A \sin(0 \alpha) + \beta B \cos(0 \beta)
+    = 0.
+\]
+In diesem Fall muss $B = 0$ gelten.
+Zusammen mit der Bedignung für $x = l$
+folgt nun
+\[
+    X^{\prime}(l)
+    =
+    - \alpha A \sin(\alpha l) + 0 \beta \cos(\beta l)
+    =
+    - \alpha A \sin(\alpha l)
+    = 0.
+\]
+
+Wiedrum muss über die $\sin$-Funktion sicher gestellt werden, dass der
+Ausdruck den Randbedingungen entspricht.
+Es folgt nun
+\[
+\begin{aligned}
+    \sin(\alpha l) &= 0 \\
+    \alpha l &= n \pi \qquad n \in \mathbb{N} \\
+    \alpha &= \frac{n \pi}{l} \qquad n \in \mathbb{N}
+\end{aligned}
+\]
+und somit
+\[
+    \mu_2 = -\alpha^{2} = -\frac{n^{2}\pi^{2}}{l^{2}}.
+\]
+
+Es ergibt sich also sowohl für einen Stab mit Enden auf konstanter Temperatur
+wie auch mit isolierten Enden
+\begin{equation}
+    \label{sturmliouville:eq:example-fourier-mu-solution}
+    \mu
+    =
+    -\frac{n^{2}\pi^{2}}{l^{2}}.
+\end{equation}
+
+%
+% Lösung von X(x), Teil: Koeffizienten a_n und b_n mittels skalarprodukt.
+%
+
+Bisher wurde über die Koeffizienten $A$ und $B$ noch nicht viel ausgesagt.
+Zunächst ist wegen vorhergehender Rechnung ersichtlich, dass es sich bei
+$A$ und $B$ nicht um einzelne Koeffizienten handelt.
+Stattdessen können die Koeffizienten für jedes $n \in \mathbb{N}$
+unterschiedlich sein.
+Die Lösung $X(x)$ wird nun umgeschrieben zu
+\[
+    X(x)
+    =
+    a_0
+    +
+    \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right)
+    +
+    \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right).
+\]
+
+Um eine eindeutige Lösung für $X(x)$ zu erhalten werden noch weitere
+Bedingungen benötigt.
+Diese sind die Startbedingungen oder $u(0, x) = X(x)$ für $t = 0$.
+Es gilt also nun die Gleichung
+\begin{equation}
+    \label{sturmliouville:eq:example-fourier-initial-conditions}
+    u(0, x)
+    =
+    a_0
+    +
+    \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right)
+    +
+    \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right)
+\end{equation}
+nach allen $a_n$ und $b_n$ aufzulösen.
+Da aber $a_n$ und $b_n$ jeweils als Faktor zu einer trigonometrischen Funktion
+gehört, von der wir wissen, dass sie orthogonal zu allen anderen
+trigonometrischen Funktionen der Lösung ist, kann direkt das Skalarprodukt
+verwendet werden um die Koeffizienten $a_n$ und $b_n$ zu bestimmen.
+Es wird also die Tatsache ausgenutzt, dass die Gleichheit in
+\eqref{sturmliouville:eq:example-fourier-initial-conditions} nach Anwendung des
+Skalarproduktes immernoch gelten muss und dass das Skalaprodukt mit einer
+Basisfunktion sämtliche Summanden auf der rechten Seite auslöscht.
+
+Zur Berechnung von $a_m$ mit $ m \in \mathbb{N} $ wird beidseitig das
+Skalarprodukt mit der Basisfunktion $ \cos\left(\frac{m \pi}{l}x\right)$
+gebildet:
+\begin{equation}
+    \label{sturmliouville:eq:dot-product-cosine}
+    \langle u(0, x), \cos\left(\frac{m \pi}{l}x\right) \rangle
+    =
+    \langle a_0
+    +
+    \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right)
+    +
+    \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right),
+    \cos\left(\frac{m \pi}{l}x\right)\rangle
+\end{equation}
+
+Bevor diese Form in die Integralform umgeschrieben werden kann, muss überlegt
+sein, welche Integralgrenzen zu verwenden sind.
+In diesem Fall haben die $\sin$ und $\cos$ Terme beispielsweise keine ganze
+Periode im Intervall $x \in [0, l]$ für ungerade $n$ und $m$.
+Um die Skalarprodukte aber korrekt zu berechnen, muss über ein ganzzahliges
+Vielfaches der Periode der trigonometrischen Funktionen integriert werden.
+Dazu werden die Integralgrenzen $-l$ und $l$ verwendet und es werden ausserdem
+neue Funktionen $\hat{u}_c(0, x)$ für die Berechnung mit Cosinus und
+$\hat{u}_s(0, x)$ für die Berechnung mit Sinus angenomen, welche $u(0, t)$
+gerade, respektive ungerade auf $[-l, l]$ fortsetzen:
+\[
+\begin{aligned}
+    \hat{u}_c(0, x)
+    &=
+    \begin{cases}
+        u(0, -x) & -l \leq x < 0
+        \\
+        u(0, x) & 0 \leq x \leq l
+    \end{cases}
+    \\
+    \hat{u}_s(0, x)
+    &=
+    \begin{cases}
+        -u(0, -x) & -l \leq x < 0
+        \\
+        u(0, x) & 0 \leq x \leq l
+    \end{cases}.
+\end{aligned}
+\]
+
+Die Konsequenz davon ist, dass nun das Resultat der Integrale um den Faktor zwei
+skalliert wurde, also gilt nun
+\[
+\begin{aligned}
+    \int_{-l}^{l}\hat{u}_c(0, x)\cos\left(\frac{m \pi}{l}x\right)dx
+    &=
+    2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx
+    \\
+    \int_{-l}^{l}\hat{u}_s(0, x)\sin\left(\frac{m \pi}{l}x\right)dx
+    &=
+    2\int_{0}^{l}u(0, x)\sin\left(\frac{m \pi}{l}x\right)dx.
+\end{aligned}
+\]
+
+Zunächst wird nun das Skalaprodukt~\eqref{sturmliouville:eq:dot-product-cosine}
+berechnet:
+\[
+\begin{aligned}
+    \int_{-l}^{l}\hat{u}_c(0, x)\cos\left(\frac{m \pi}{l}x\right)dx
+    =&
+    \int_{-l}^{l} \left[a_0
+    +
+    \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right)
+    +
+    \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right)\right]
+    \cos\left(\frac{m \pi}{l}x\right) dx
+    \\
+    2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx
+    =&
+    a_0 \int_{-l}^{l}\cos\left(\frac{m \pi}{l}x\right) dx
+    +
+    \sum_{n = 1}^{\infty}\left[a_n\int_{-l}^{l}\cos\left(\frac{n\pi}{l}x\right)
+        \cos\left(\frac{m \pi}{l}x\right)dx\right]
+    \\
+    &+
+    \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right)
+        \cos\left(\frac{m \pi}{l}x\right)dx\right].
+\end{aligned}
+\]
+
+Betrachtet man nun die Summanden auf der rechten Seite stellt man fest, dass
+nahezu alle Terme verschwinden, denn
+\[
+    \int_{-l}^{l}\cos\left(\frac{m \pi}{l}x\right) dx
+    =
+    0,
+\]
+da hier über ein ganzzahliges Vielfaches der Periode integriert wird,
+\[
+    \int_{-l}^{l}\cos\left(\frac{n\pi}{l}x\right)
+    \cos\left(\frac{m \pi}{l}x\right)dx
+    =
+    0
+\]
+für $m\neq n$, da Cosinus-Funktionen mit verschiedenen Kreisfrequenzen
+orthogonal zueinander stehen und
+\[
+    \int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right)
+        \cos\left(\frac{m \pi}{l}x\right)dx
+    =
+    0
+\]
+da Sinus- und Cosinus-Funktionen ebenfalls orthogonal zueinander sind.
+
+Es bleibt also lediglich der Summand für $a_m$ stehen, was die Gleichung zu
+\[
+    2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx
+    =
+    a_m\int_{-l}^{l}\cos^2\left(\frac{m\pi}{l}x\right)dx
+\]
+vereinfacht. Im nächsten Schritt wird nun das Integral auf der rechten Seite
+berechnet und dann nach $a_m$ aufgelöst. Am einnfachsten geht dies, wenn zuerst
+mit $u = \frac{m \pi}{l}x$ substituiert wird:
+\[
+    \begin{aligned}
+    2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx
+    &=
+    a_m\frac{l}{m\pi}\int_{-m\pi}^{m\pi}\cos^2\left(u\right)du
+    \\
+    &=
+    a_m\frac{l}{m\pi}\left[\frac{u}{2} + 
+    \frac{\sin\left(2u\right)}{4}\right]_{u=-m\pi}^{m\pi}
+    \\
+    &=
+    a_m\frac{l}{m\pi}\biggl(\frac{m\pi}{2} + 
+    \underbrace{\frac{\sin\left(2m\pi\right)}{4}}_{\displaystyle = 0} - 
+    \frac{-m\pi}{2} -
+    \underbrace{\frac{\sin\left(-2m\pi\right)}{4}}_{\displaystyle = 0}\biggr)
+    \\
+    &=
+    a_m l
+    \\
+    a_m
+    &=
+    \frac{2}{l} \int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx
+    \end{aligned}
+\]
+
+Analog dazu kann durch das Bilden des Skalarproduktes mit 
+$ \sin\left(\frac{m \pi}{l}x\right) $ gezeigt werden, dass
+\[
+    b_m
+    =
+    \frac{2}{l} \int_{0}^{l}u(0, x)\sin\left(\frac{m \pi}{l}x\right)dx
+\]
+gilt.
+
+Etwas anders ist es allerdings bei $a_0$.
+Wie der Name bereits suggeriert, handelt es sich hierbei um den Koeffizienten
+zur Basisfunktion $\cos\left(\frac{0 \pi}{l}x\right)$ beziehungsweise der
+konstanten Funktion $1$.
+Um einen Ausdruck für $a_0$ zu erhalten, wird wiederum auf beiden Seiten
+der Gleichung~\eqref{sturmliouville:eq:example-fourier-initial-conditions} das
+Skalarprodukt mit der konstanten Basisfunktion $1$ gebildet:
+\[
+\begin{aligned}
+    \int_{-l}^{l}\hat{u}_c(0, x)dx
+    &=
+    \int_{-l}^{l} a_0
+    +
+    \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right)
+    +
+    \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right)dx
+    \\
+    2\int_{0}^{l}u(0, x)dx
+    &=
+    a_0 \int_{-l}^{l}dx
+    +
+    \sum_{n = 1}^{\infty}\left[a_n\int_{-l}^{l}\cos\left(\frac{n\pi}{l}x\right)
+        dx\right] +
+    \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right)
+        dx\right].
+\end{aligned}
+\]
+
+Hier fallen nun alle Terme, die $\sin$ oder $\cos$ beinhalten weg, da jeweils
+über ein Vielfaches der Periode integriert wird.
+Es bleibt also noch
+\[
+    2\int_{0}^{l}u(0, x)dx
+    =
+    a_0 \int_{-l}^{l}dx
+\]
+, was sich wie folgt nach $a_0$ auflösen lässt:
+\[
+\begin{aligned}
+    2\int_{0}^{l}u(0, x)dx
+    &=
+    a_0 \int_{-l}^{l}dx
+    \\
+    &=
+    a_0 \left[x\right]_{x=-l}^{l}
+    \\
+    &=
+    a_0(l - (-l))
+    \\
+    &=
+    a_0 \cdot 2l
+    \\
+    a_0
+    &=
+    \frac{1}{l} \int_{0}^{l}u(0, x)dx
+\end{aligned}
+\]
+
+%
+% Lösung von T(t) 
+%
+
+\subsubsection{Lösung der Differentialgleichung in $t$}
+Zuletzt wird die zweite Gleichung der 
+Separation~\eqref{sturmliouville:eq:example-fourier-separated-t} betrachtet.
+Diese wird über das charakteristische Polynom
+\[
+    \lambda - \kappa \mu
+    =
+    0
+\]
+gelöst.
+
+Es ist direkt ersichtlich, dass $\lambda = \kappa \mu$ gelten muss, was zur
+Lösung
+\[
+    T(t)
+    =
+    e^{\kappa \mu t}
+\]
+führt und mit dem Resultat~\eqref{sturmliouville:eq:example-fourier-mu-solution}
+\[
+    T(t)
+    =
+    e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t}
+\]
+ergibt.
+
+Dieses Resultat kann nun mit allen vorhergehenden Resultaten zusammengesetzt
+werden um die vollständige Lösung für das Stab-Problem zu erhalten.
+
+\subsubsection{Lösung für einen Stab mit Enden auf konstanter Temperatur}
+\[
+\begin{aligned}
+    u(t,x)
+    &=
+    \sum_{n=1}^{\infty}b_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t}
+    \sin\left(\frac{n\pi}{l}x\right)
+    \\
+    b_{n}
+    &=
+    \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx
+\end{aligned}
+\]
+
+\subsubsection{Lösung für einen Stab mit isolierten Enden}
+\[
+\begin{aligned}
+    u(t,x)
+    &=
+    a_{0} + \sum_{n=1}^{\infty}a_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t}
+    \cos\left(\frac{n\pi}{l}x\right)
+    \\
+    a_{0}
+    &=
+    \frac{1}{l}\int_{0}^{l}u(0,x) dx
+    \\
+    a_{n}
+    &=
+    \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx
+\end{aligned}
+\]
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 @@
+%% Creator: Matplotlib, PGF backend
+%%
+%% To include the figure in your LaTeX document, write
+%%   \input{<filename>.pgf}
+%%
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+%%   \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}
+%%
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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
+%%
+\begingroup%
+\makeatletter%
+\begin{pgfpicture}%
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+\begin{pgfscope}%
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+}%
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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
+++ b/buch/papers/zeta/images/youtube_screenshot.png
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
+++ b/buch/papers/zeta/images/zeta_re_-1_plot.pgf
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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
+++ b/buch/papers/zeta/images/zeta_re_0.5_paper.pgf
@@ -0,0 +1,1137 @@
+%% 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
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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
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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
+%%
+%% 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}{-}
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+%%
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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})
+	-- ({0.9625*\dx},{-0.7553*\dy})
+	-- ({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})
+	-- ({0.8963*\dx},{-0.7626*\dy})
+	-- ({0.8850*\dx},{-0.7632*\dy})
+	-- ({0.8738*\dx},{-0.7637*\dy})
+	-- ({0.8624*\dx},{-0.7640*\dy})
+	-- ({0.8510*\dx},{-0.7642*\dy})
+	-- ({0.8396*\dx},{-0.7642*\dy})
+	-- ({0.8281*\dx},{-0.7640*\dy})
+	-- ({0.8166*\dx},{-0.7636*\dy})
+	-- ({0.8050*\dx},{-0.7631*\dy})
+	-- ({0.7934*\dx},{-0.7624*\dy})
+	-- ({0.7818*\dx},{-0.7615*\dy})
+	-- ({0.7702*\dx},{-0.7605*\dy})
+	-- ({0.7585*\dx},{-0.7593*\dy})
+	-- ({0.7468*\dx},{-0.7579*\dy})
+	-- ({0.7350*\dx},{-0.7563*\dy})
+	-- ({0.7233*\dx},{-0.7545*\dy})
+	-- ({0.7115*\dx},{-0.7526*\dy})
+	-- ({0.6998*\dx},{-0.7504*\dy})
+	-- ({0.6880*\dx},{-0.7481*\dy})
+	-- ({0.6762*\dx},{-0.7456*\dy})
+	-- ({0.6644*\dx},{-0.7429*\dy})
+	-- ({0.6526*\dx},{-0.7401*\dy})
+	-- ({0.6407*\dx},{-0.7370*\dy})
+	-- ({0.6289*\dx},{-0.7337*\dy})
+	-- ({0.6171*\dx},{-0.7303*\dy})
+	-- ({0.6054*\dx},{-0.7267*\dy})
+	-- ({0.5936*\dx},{-0.7229*\dy})
+	-- ({0.5818*\dx},{-0.7188*\dy})
+	-- ({0.5701*\dx},{-0.7146*\dy})
+	-- ({0.5583*\dx},{-0.7102*\dy})
+	-- ({0.5466*\dx},{-0.7056*\dy})
+	-- ({0.5350*\dx},{-0.7008*\dy})
+	-- ({0.5233*\dx},{-0.6959*\dy})
+	-- ({0.5117*\dx},{-0.6907*\dy})
+	-- ({0.5002*\dx},{-0.6853*\dy})
+	-- ({0.4886*\dx},{-0.6797*\dy})
+	-- ({0.4772*\dx},{-0.6740*\dy})
+	-- ({0.4657*\dx},{-0.6680*\dy})
+	-- ({0.4543*\dx},{-0.6618*\dy})
+	-- ({0.4430*\dx},{-0.6555*\dy})
+	-- ({0.4317*\dx},{-0.6489*\dy})
+	-- ({0.4205*\dx},{-0.6422*\dy})
+	-- ({0.4094*\dx},{-0.6352*\dy})
+	-- ({0.3983*\dx},{-0.6281*\dy})
+	-- ({0.3873*\dx},{-0.6208*\dy})
+	-- ({0.3763*\dx},{-0.6132*\dy})
+	-- ({0.3655*\dx},{-0.6055*\dy})
+	-- ({0.3547*\dx},{-0.5976*\dy})
+	-- ({0.3440*\dx},{-0.5894*\dy})
+	-- ({0.3334*\dx},{-0.5811*\dy})
+	-- ({0.3229*\dx},{-0.5726*\dy})
+	-- ({0.3125*\dx},{-0.5639*\dy})
+	-- ({0.3022*\dx},{-0.5550*\dy})
+	-- ({0.2920*\dx},{-0.5459*\dy})
+	-- ({0.2818*\dx},{-0.5367*\dy})
+	-- ({0.2718*\dx},{-0.5272*\dy})
+	-- ({0.2620*\dx},{-0.5175*\dy})
+	-- ({0.2522*\dx},{-0.5077*\dy})
+	-- ({0.2425*\dx},{-0.4977*\dy})
+	-- ({0.2330*\dx},{-0.4875*\dy})
+	-- ({0.2236*\dx},{-0.4771*\dy})
+	-- ({0.2144*\dx},{-0.4665*\dy})
+	-- ({0.2052*\dx},{-0.4557*\dy})
+	-- ({0.1962*\dx},{-0.4448*\dy})
+	-- ({0.1874*\dx},{-0.4337*\dy})
+	-- ({0.1787*\dx},{-0.4224*\dy})
+	-- ({0.1701*\dx},{-0.4109*\dy})
+	-- ({0.1617*\dx},{-0.3992*\dy})
+	-- ({0.1535*\dx},{-0.3874*\dy})
+	-- ({0.1454*\dx},{-0.3754*\dy})
+	-- ({0.1375*\dx},{-0.3633*\dy})
+	-- ({0.1297*\dx},{-0.3509*\dy})
+	-- ({0.1221*\dx},{-0.3384*\dy})
+	-- ({0.1147*\dx},{-0.3258*\dy})
+	-- ({0.1074*\dx},{-0.3130*\dy})
+	-- ({0.1004*\dx},{-0.3000*\dy})
+	-- ({0.0935*\dx},{-0.2869*\dy})
+	-- ({0.0868*\dx},{-0.2736*\dy})
+	-- ({0.0803*\dx},{-0.2602*\dy})
+	-- ({0.0740*\dx},{-0.2466*\dy})
+	-- ({0.0679*\dx},{-0.2329*\dy})
+	-- ({0.0620*\dx},{-0.2190*\dy})
+	-- ({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
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+%%
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+%%
+%% and, on pdftex
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+%%
+%% 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
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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%
+\pgfsetmiterjoin%
+\definecolor{currentfill}{rgb}{1.000000,1.000000,1.000000}%
+\pgfsetfillcolor{currentfill}%
+\pgfsetlinewidth{0.000000pt}%
+\definecolor{currentstroke}{rgb}{1.000000,1.000000,1.000000}%
+\pgfsetstrokecolor{currentstroke}%
+\pgfsetdash{}{0pt}%
+\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}%
+\pgfpathlineto{\pgfqpoint{6.400000in}{0.000000in}}%
+\pgfpathlineto{\pgfqpoint{6.400000in}{4.800000in}}%
+\pgfpathlineto{\pgfqpoint{0.000000in}{4.800000in}}%
+\pgfpathclose%
+\pgfusepath{fill}%
+\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{2.588156in}{0.528000in}}%
+\pgfpathlineto{\pgfqpoint{3.971844in}{0.528000in}}%
+\pgfpathlineto{\pgfqpoint{3.971844in}{4.224000in}}%
+\pgfpathlineto{\pgfqpoint{2.588156in}{4.224000in}}%
+\pgfpathclose%
+\pgfusepath{fill}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\pgfpathrectangle{\pgfqpoint{2.588156in}{0.528000in}}{\pgfqpoint{1.383688in}{3.696000in}}%
+\pgfusepath{clip}%
+\pgfsys@transformshift{2.588156in}{0.528000in}%
+\pgftext[left,bottom]{\includegraphics[interpolate=true,width=1.390000in,height=3.700000in]{zeta_color_plot-img0.png}}%
+\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{2.588156in}{0.528000in}%
+\pgfsys@useobject{currentmarker}{}%
+\end{pgfscope}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{textcolor}%
+\pgfsetfillcolor{textcolor}%
+\pgftext[x=2.588156in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {-10}\)}%
+\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{3.050619in}{0.528000in}%
+\pgfsys@useobject{currentmarker}{}%
+\end{pgfscope}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{textcolor}%
+\pgfsetfillcolor{textcolor}%
+\pgftext[x=3.050619in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {-5}\)}%
+\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}}%
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+\pgfusepath{stroke,fill}%
+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{3.513081in}{0.528000in}%
+\pgfsys@useobject{currentmarker}{}%
+\end{pgfscope}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{textcolor}%
+\pgfsetfillcolor{textcolor}%
+\pgftext[x=3.513081in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {0}\)}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{textcolor}%
+\pgfsetfillcolor{textcolor}%
+\pgftext[x=3.280000in,y=0.276457in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle \Re\)}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\pgfsetbuttcap%
+\pgfsetroundjoin%
+\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}%
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+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{2.588156in}{0.528000in}%
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+\end{pgfscope}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{textcolor}%
+\pgfsetfillcolor{textcolor}%
+\pgftext[x=2.281054in, y=0.489420in, left, base]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {-20}\)}%
+\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}%
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+\pgfusepath{stroke,fill}%
+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{2.588156in}{0.990462in}%
+\pgfsys@useobject{currentmarker}{}%
+\end{pgfscope}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfsetfillcolor{textcolor}%
+\pgftext[x=2.281054in, y=0.951882in, left, base]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {-15}\)}%
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+\begin{pgfscope}%
+\pgfsetbuttcap%
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+\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfsetlinewidth{0.803000pt}%
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+\pgfusepath{stroke,fill}%
+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{2.588156in}{1.452925in}%
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+\end{pgfscope}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
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+\begin{pgfscope}%
+\pgfsetbuttcap%
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+\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfsetlinewidth{0.803000pt}%
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+\pgfusepath{stroke,fill}%
+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{2.588156in}{1.915387in}%
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+\end{pgfscope}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfsetbuttcap%
+\pgfsetroundjoin%
+\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfsetlinewidth{0.803000pt}%
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+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{2.588156in}{2.377850in}%
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+\pgfsetbuttcap%
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+\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}%
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+\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
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+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{2.588156in}{2.840312in}%
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+\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfsetlinewidth{0.803000pt}%
+\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfpathmoveto{\pgfqpoint{-0.000000in}{0.000000in}}%
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+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{2.588156in}{3.302775in}%
+\pgfsys@useobject{currentmarker}{}%
+\end{pgfscope}%
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+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfsetbuttcap%
+\pgfsetroundjoin%
+\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}%
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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}