diff --git a/zusammenfassung/analysis/AnalysisZF.aux b/zusammenfassung/analysis/AnalysisZF.aux
new file mode 100644
index 0000000..8c90ba8
--- /dev/null
+++ b/zusammenfassung/analysis/AnalysisZF.aux
@@ -0,0 +1,637 @@
+\relax 
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+\providecommand\babel@aux[2]{}
+\@nameuse{bbl@beforestart}
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diff --git a/zusammenfassung/analysis/AnalysisZF.fdb_latexmk b/zusammenfassung/analysis/AnalysisZF.fdb_latexmk
new file mode 100644
index 0000000..b43a602
--- /dev/null
+++ b/zusammenfassung/analysis/AnalysisZF.fdb_latexmk
@@ -0,0 +1,271 @@
+# Fdb version 4
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+  
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+
+\begin{document}
+
+\setcounter{secnumdepth}{0} %no enumeration of sections
+\begin{multicols*}{3}
+    \begin{center}
+        \Large{Analysis 1 \& 2} \\
+        \tiny{von Jirayu Ruh, \href{mailto:jirruh@ethz.ch}{jirruh@ethz.ch}}
+    \end{center} 
+
+
+    \section{Grundlagen}
+    \subsection{Logik}
+
+    \begin{center}
+        \begin{tabular}{ c c c c c c c c} \toprule
+                & \hspace*{-10pt}     & \hspace*{-10pt} Negation & \hspace*{-10pt} AND         & \hspace*{-10pt} OR         & \hspace*{-10pt} Implikation       & \hspace*{-10pt} Äquivalenz            \\
+            $A$ & \hspace*{-10pt} $B$ & \hspace*{-10pt} $\neg A$ & \hspace*{-10pt} $A \land B$ & \hspace*{-10pt} $A \lor B$ & \hspace*{-10pt} $A \rightarrow B$ & \hspace*{-10pt} $A \leftrightarrow B$ \\ \midrule
+            w   & \hspace*{-10pt} w   & \hspace*{-10pt} f        & \hspace*{-10pt} w           & \hspace*{-10pt} w          & \hspace*{-10pt} w                 & \hspace*{-10pt} w                     \\
+            w   & \hspace*{-10pt} f   & \hspace*{-10pt} f        & \hspace*{-10pt} f           & \hspace*{-10pt} w          & \hspace*{-10pt} f                 & \hspace*{-10pt} f                     \\
+            f   & \hspace*{-10pt} w   & \hspace*{-10pt} w        & \hspace*{-10pt} f           & \hspace*{-10pt} w          & \hspace*{-10pt} w                 & \hspace*{-10pt} f                     \\
+            f   & \hspace*{-10pt} f   & \hspace*{-10pt} w        & \hspace*{-10pt} f           & \hspace*{-10pt} f          & \hspace*{-10pt} w                 & \hspace*{-10pt} w                     \\ \bottomrule
+        \end{tabular}
+    \end{center}
+
+    i) Wahre Implikation: $A \Rightarrow B$ (''A ist hinreichend für B'').
+
+    ii) Wahre Äquivalenz: $A \Leftrightarrow B$ (''A gilt genau dann, wenn B gilt'').
+
+    \begin{center}
+        Negation der Implikation: \quad \eqbox{$\neg(A \rightarrow B) \Leftrightarrow A \land \neg B$}
+    \end{center}
+
+    \subsubsection{Kontraposition}
+
+    Falls $A \Rightarrow B$, so gilt auch $\neg B \Rightarrow \neg A$ (''B ist notwendig für A'').
+
+
+    \subsubsection{Indirekter Beweis}
+
+    Zum Beweis der Aussage $A \Rightarrow B$ genügt es die Aussage $\neg B \Rightarrow \neg A$ zu zeigen oder die Annahme $A \land \neg B$ zum Widerspruch zu führen.
+
+
+    \subsubsection{Prinzip der vollständigen Induktion}
+
+    Sei $A(n)$ eine Aussage mit $n \in \N$.
+
+    \begin{tabular}{r p{0.28\textwidth}} \toprule
+        i)   & \hspace*{-10pt} Induktions-Verankerung: Zeige, dass $A(1)$ gilt.                                   \\
+        ii)  & \hspace*{-10pt} Induktions-Annahme: Annahme, dass $A(n)$ gilt.                                     \\
+        iii) & \hspace*{-10pt} Induktionsschritt: Beweise, dass $A(n+1)$ gilt unter der Annahme, dass $A_n$ gilt.
+        Achtung, nichts unbewiesenes gleichsetzen!                                                                \\ \bottomrule
+    \end{tabular}
+
+
+    \subsection{Mengenlehre}
+
+    Eine Menge wird oft bestimmt durch eine Bedingung $A(b)$ wobei $b \in X$:
+
+    \begin{center}
+        \eqbox{$Y= \{b \in X ; A(b) \}$}
+    \end{center}
+
+    \begin{center}
+        \begin{tabular}{r l l} \toprule
+            Notation        & Definition                                                        &                                                  \\ \midrule
+            $\{\dots\}$     & \multicolumn{2}{l}{Set: Sammlung von ungeordneten Elemente}                                                          \\
+            $(\dots)$       & \multicolumn{2}{l}{Tupel: Sammlung von geordneten Elementen}                                                         \\
+            $A \cup B$      & Vereinigungsmenge                                                 & $ := \{x \in \R;(x \in A) \lor (x \in B)\}$      \\
+            $A \cap B$      & Schnittmenge                                                      & $ := \{x \in \R;(x \in A) \land (x \in B)\}$     \\
+            $A \setminus B$ & Differenzmenge                                                    & $ := \{x \in \R; (x \in A) \land (x \notin B)\}$ \\
+            $A^C$           & \multicolumn{2}{l}{Komplement, alle Elemente die nicht in A sind}                                                    \\
+            $A \subset B$   & \multicolumn{2}{l}{A ist eine Teilmenge (oder gleich) von B}                                                         \\
+            $\emptyset$     & Leeres Set                                                        &                                                  \\\bottomrule
+        \end{tabular}
+    \end{center}
+
+
+    \subsubsection{Quantoren}
+
+    \begin{center}
+        \begin{tabular}{r l } \toprule
+            Quantor            & Beschreibung                              \\ \midrule
+            $\forall x, A(x)$  & Für alle x gilt $A(x)$                    \\
+            $\exists x, A(x)$  & Es existiert min. ein x, wo $A(x)$ gilt.  \\
+            $\exists! x, A(x)$ & Es existiert genau ein x, wo $A(x)$ gilt. \\
+            $\nexists x, A(x)$ & Es existiert kein x, wo $A(x)$ gilt.      \\ \bottomrule
+        \end{tabular}
+    \end{center}
+
+    Negation: \eqbox{$\neg (\forall x, A(x)) \Leftrightarrow \exists x, \neg A(x) \quad \neg (\exists x, A(x)) \Leftrightarrow \forall x, \neg A(x)$}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Funktionen (Abbildungen)}
+
+    Eine Abbildung $f$ mit Definitionsbereich $X$ und Bild-/Wertebereich $Y$:
+
+    \begin{center}
+        \eqbox{$f: X \to Y, \quad x \mapsto f(x)$}
+    \end{center}
+
+    \begin{center}
+        \begin{tabular}{r l} \toprule
+            Definition               & Beschreibung                                  \\ \midrule
+            Urbild von $B \subset Y$ & $f^{-1}(B) := \{x \in X: f(x) \in B\}$        \\
+            Identität                & $id_X : X \to X, \quad x \mapsto x = id_X(x)$ \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \subsubsection{Komposition}
+
+    Sei $f: X \to Y$ und $g: Y \to Z$. Dann ist die Komposition von $f$ und $g$:
+
+    \begin{center}
+        \eqbox{$F:= g \circ f: X \to Z, \quad x \mapsto g(f(x))$}
+    \end{center}
+
+    Die Komposition ist Assoziativ: $(h \circ g) \circ f = h \circ (g \circ f)$
+
+
+    \begin{center}
+        \begin{minipage}{0.5\linewidth}
+            \subsubsection{Surjektiv}
+
+            $f$ heisst surjektiv falls jedes $y \in Y$ \emph{mindestens} ein Urbild hat, d.h.:
+            \begin{center}
+                $\forall y \in Y, \, \exists x \in X :  f(x) = y$
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.49\linewidth}
+            \begin{center}
+                \includegraphics[width=0.75\linewidth]{Bilder/Surjektiv.JPG}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+
+    \begin{center}
+        \begin{minipage}{0.5\linewidth}
+            \subsubsection{Injektiv}
+
+            $f$ heisst injektiv falls jedes $y \in Y$ \emph{höchstens} ein Urbild hat, d.h.:
+
+            \begin{center}
+                $\forall x_1, x_2 \in X : f(x_1) = f(x_2) \Rightarrow x_1 = x_2$
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.49\linewidth}
+            \begin{center}
+                \includegraphics[width=0.75\linewidth]{Bilder/Injektiv.JPG}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \subsubsection{Bijektiv und Umkehrabbildung}
+
+    $f$ heisst bijektiv, falls jedes $y \in Y$ \emph{genau} ein Urbild hat, d.h. $f$ ist surjektiv und injektiv.
+
+    Ist $f$ bijektiv, dann kann man eine \emph{Umkehrabbildung} $f^{-1}$ einführen:
+
+    \begin{center}
+        \eqbox{$f^{-1}: Y \to X, \quad y \mapsto f(y)$}
+    \end{center}
+
+
+    \subsection{Reelle Zahlen}
+
+    \begin{center}
+        \begin{tabular}{r l} \toprule
+            Natürliche Zahlen             & $\mathbb{N} = \{1, 2, 3, \dots\}, \, \mathbb{N}_0 = \{0, 1, 2, 3, \dots\}$ \\
+            Ganze Zahlen                  & $\mathbb{Z} = \{\dots, -1, 0, 1, \dots\}$                                  \\
+            Rationale Zahlen              & $\mathbb{Q} = \{\frac{p}{q}; p \in \mathbb{Z} \land q \in \mathbb{N}\}$    \\
+            Irrationale Zahlen            & $\R \setminus \mathbb{Q}$                                                  \\
+            Reelle Zahlen                 & $\R = \mathbb{Q} + \R \setminus \mathbb{Q}$                                \\ \midrule
+            $[a,b]$                       & \hspace*{-10pt} $ := \{x \in \R; a \leq x \leq b\}$                        \\
+            $]a,b[ \Leftrightarrow (a,b)$ & \hspace*{-10pt} $:= \{x \in \R; a < x < b\}$                               \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \subsubsection{Vollständigkeitsaxiom}
+
+    $\R$ ist \emph{Ordnungsvollständig}, dass heisst: \eqbox{$\forall a, b \in \R \,\, \exists c \in \R: \, a \leq c \leq b$}
+
+    \subsubsection{Dreiecksungleichung}
+
+    Es gilt für alle $x,y \in \R$: \eqboxf{$| x + y | \leq | x | + | y |$}
+
+    \subsubsection{Archimedisches Prinzip}
+
+    Zu jeder Zahl $0 < b \in \R$ gibt es ein $n \in \mathbb{N}$ mit $b < n$. \medskip
+
+    Daraus folgt: $\infty$ und $-\infty$ ist keine reelle Zahl.
+    \vfill\null
+    \columnbreak
+
+
+    \subsubsection{Supremum und Infimum}
+
+
+    \begin{center}
+        \begin{minipage}{0.48\linewidth}
+            Eine Menge $A \subset \R$ heisst nach oben beschränkt, falls gilt
+
+            \begin{center}
+                $\exists b \in \R \, \, \forall a \in A: a \leq b$
+            \end{center}
+
+            wobei $b$ eine obere Schranke genannt wird, die kleinste obere Schranke ist das \textbf{Supremum}.
+
+            \begin{center}
+                $\sup\limits_{x \in \R} A \Leftrightarrow \sup\{A; x \in \R\}$
+            \end{center}
+
+            Wird das Supremum angenommen in $A$, dann ist es das \textbf{Maximum}.
+            \vfill\null
+        \end{minipage}
+        \,\vline\,
+        \begin{minipage}{0.48\linewidth}
+            Eine Menge $A \subset \R$ heisst nach unten beschränkt, falls gilt
+
+            \begin{center}
+                $\exists b \in \R \, \, \forall a \in A: a \geq b$
+            \end{center}
+
+            wobei $b$ eine untere Schranke genannt wird, die kleinste untere Schranke ist das \textbf{Infimum}.
+
+            \begin{center}
+                $\inf\limits_{x \in \R} A \Leftrightarrow \inf\{A; x \in \R\}$
+            \end{center}
+
+            Wird das Infimum angenommen in $A$, dann ist es das \textbf{Minimum}.
+            \vfill\null
+        \end{minipage}
+    \end{center}
+
+
+    \subsection{Potenzen und Wurzel}
+
+    Sei $n,m \in \N$. Die reelle Wurzel ist definiert auf $\R^+$.
+
+    \begin{center}
+        \begin{minipage}{0.48\linewidth}
+            \begin{center}
+                \renewcommand{\arraystretch}{1.35}
+                \begin{tabular}{r l}
+                    $a^n \cdot a^m$    & \hspace*{-10pt}$= a^{n + m}$
+                    \\
+                    $\dfrac{a^n}{a^m}$ & \hspace*{-10pt}$= a^{n - m}$                   \\
+                    $a^n \cdot b^n$    & \hspace*{-10pt}$= (a \cdot b)^n$               \\
+                    $\dfrac{a^n}{b^n}$ & \hspace*{-10pt}$= \left(\dfrac{a}{b}\right)^n$ \\
+                    $(a^n)^m$          & \hspace*{-10pt}$= a^{n \cdot m}$               \\
+                    $b^0$              & \hspace*{-10pt}$= 1$                           \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+        \,\vline\,
+        \begin{minipage}{0.48\linewidth}
+            \begin{center}
+                \renewcommand{\arraystretch}{1.35}
+                \begin{tabular}{r l}
+                    $b^{\frac{n}{m}}$                  & \hspace*{-10pt}$= \sqrt[m]{b^n}$          \\
+                    $b^{-1}$                           & \hspace*{-10pt}$= \dfrac{1}{b^n}$         \\
+                    $\sqrt[n]{a} \cdot \sqrt[n]{b}$    & \hspace*{-10pt}$= \sqrt[n]{a \cdot b}$
+                    \\
+                    $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ & \hspace*{-10pt}$= \sqrt[n]{\dfrac{a}{b}}$ \\
+                    $\sqrt[n]{\sqrt[m]{a}}$            & \hspace*{-10pt}$= \sqrt[n \cdot m]{a}$    \\
+                    $\sqrt[n]{1}$                      & \hspace*{-10pt}$= 1$                      \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \textbf{Achtung}: Beim Wurzel ziehen, immer $\pm$ vor der Wurzel!
+
+
+    \subsection{Logarithmus}
+
+    Der Logarithmus ist definiert auf $\R^+ \setminus {0}$. Eigenschaften:
+
+    \begin{center}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \renewcommand{\arraystretch}{1.35}
+                \begin{tabular}{r l}
+                    $\log(a \cdot b)$               & \hspace*{-10pt}$= \log(a) + \log(b)$
+                    \\
+                    $\log\left(\dfrac{a}{b}\right)$ & \hspace*{-10pt}$= \log(a) - \log(b)$ \\
+                    $\log(a^n)$                     & \hspace*{-10pt}$= n \cdot \log(a)$   \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+        \,\vline\,
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \renewcommand{\arraystretch}{1.35}
+                \begin{tabular}{r l}
+                    $e^{\log(a)}$ & \hspace*{-10pt}$= a$ \\
+                    $\log(e)$     & \hspace*{-10pt}$= 1$ \\
+                    $\log(1)$     & \hspace*{-10pt}$= 0$ \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    Bemerkung: $x = e^n \Leftrightarrow \log(x) = n$
+
+
+    \subsection{Die Exponentialfunktion ($\exp(x) = e^x$)}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.35}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \begin{tabular}{r l}
+                    $e^{z + w}$    & \hspace*{-10pt}$= e^z \cdot e^w$  \\
+                    $e^{-x}$       & \hspace*{-10pt}$= \dfrac{1}{e^x}$ \\
+                    $\exp^{-1}(y)$ & \hspace*{-10pt}$= \log(y)$        \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+        \,\vline\,
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \renewcommand{\arraystretch}{1.35}
+                \begin{tabular}{r l}
+                    $\lim\limits_{x \to \infty} e^x$  & \hspace*{-10pt}$= \infty$ \\
+                    $e^0$                             & \hspace*{-10pt}$= 1$      \\
+                    $\lim\limits_{x \to -\infty} e^x$ & \hspace*{-10pt}$= 0$      \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \subsubsection{Trigonometrische- und Hyperbelfunktionen}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{2.5}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \begin{tabular}{r l}
+                    $\sin(z)$ & \hspace*{-10pt}$= \dfrac{e^{iz} - e^{-iz}}{2i}$ \\
+                    $\cos(z)$ & \hspace*{-10pt}$= \dfrac{e^{iz} + e^{-iz}}{2}$  \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \begin{tabular}{r l}
+                    $\sinh(z)$ & \hspace*{-10pt}$= \dfrac{e^{z} - e^{-z}}{2}$ \\
+                    $\cosh(z)$ & \hspace*{-10pt}$= \dfrac{e^{z} + e^{-z}}{2}$ \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Vektoren}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \begin{tabular}{r l} \toprule
+            Euklidischer Raum    & $\R^n = \{\vect{x} = (x_1, \dots, x_n); x_i \in \R\}$                                   \\
+            Addition             & $\vect{x} + \vect{y} = (x_1 + y_1, \dots, x_n + y_n), \, x_i, y_i \in \R^n$             \\
+            Skalarmultiplikation & $\lambda \vect{x} = (\lambda x_1, \dots, \lambda x_n), \, x _i\in \R^n, \lambda \in \R$ \\
+            Skalarprodukt        & $\langle \vect{x}, \vect{y} \rangle = x_1 y_1 + \dots + x_n y_n$                        \\
+            Euklidische Norm     & $|| x || := \sqrt{\displaystyle\sum^n_{i = 1} x_i^2}$                                   \\ \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \subsubsection{Cauchy-Schwarz}
+
+    Für alle $\vect{x}, \vect{y} \in \R^n$ gilt:
+
+    \begin{center}
+        \eqbox{$| \vect{x} \cdot \vect{y} | \leq || \vect{x} || \cdot || \vect{y} ||$}
+    \end{center}
+
+
+    \subsection{Komplexe Zahlen}
+
+    \begin{center}
+        \eqbox{$\mathbb{C} := \{a + i b: a,b \in \R\}$} wobei \eqboxf{$i = \sqrt{-1}$}
+    \end{center}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \begin{tabular}{r l} \toprule
+            Realteil             & $\Re{z} = a = \dfrac{z + \overline{z}}{2}$                                                                                    \\
+            Imaginärteil         & $\Im{z} = b = \dfrac{z - \overline{z}}{2i}$                                                                                   \\ \midrule
+            Komplexe Konjugation & $\overline{z} = a - ib$                                                                                                       \\
+                                 & $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$, \, $\overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2}$ \\ \midrule
+            Addition             & $z_1 + z_2 = (a_1 + a_2) + i(b_1 + b_2)$                                                                                      \\
+            Multiplikation       & $z_1 \cdot z_2 = (a_1 a_2 - b_1 b_2) + i(a_2 b_1 + a_1 b_2)$                                                                  \\
+            Division             & $\dfrac{z_1}{z_2} = \dfrac{z_1 \cdot \overline{z_2}}{z_2 \cdot \overline{z_2}}$                                               \\ \midrule
+            Absolutbetrag        & $| z | = \sqrt{z \overline{z}} = \sqrt{\Re{z}^2 + \Im{z}^2}$                                                                  \\
+                                 & $|z_1 \cdot z_2| = |z_1| \cdot |z_2|$                                                                                         \\
+            Phase                & $\varphi = \arctan \left( \dfrac{\Im{z}}{\Re{z}} \right)$                                                                     \\ \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \subsubsection{Eulersche Formel und Eulers Identität}
+
+    \begin{center}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \eqboxf{$e^{i\varphi} = \cos\varphi + i \sin\varphi$}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \eqbox{$e^{i \pi} = -1$} \qquad \eqbox{$e^{2 \pi i} = 1$}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \subsubsection{Polarform}
+
+    In der Polardarstellung $z = | z | e^{i\varphi}$ gelten folgende Rechenregel:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{r l} \toprule
+            Realteil            & $\Re{z} = \cos(\varphi)$                                                                                               \\
+            Imaginärteil        & $\Im{z} = \sin(\varphi)$                                                                                               \\
+            \midrule
+            Komplex Konjugation & \hspace*{-10pt} $\overline{z} = | z | e^{-i\varphi} = |z| \cdot (\cos\varphi - i\sin\varphi)$                          \\ \midrule
+            Multiplikation      & \hspace*{-10pt} $z_1 \cdot z_2 = |z_1| \cdot |z_2| e^{i(\varphi_1+\varphi_2)}$                                         \\
+            Division            & \hspace*{-10pt} $\dfrac{z_1}{z_2} = \dfrac{|z_1|}{|z_2|} \cdot e^{i(\varphi_1 - \varphi_2)}$                           \\
+            \midrule
+            Potenzieren         & \hspace*{-10pt} $(|z|e^{i\varphi})^n = |z|^n \cdot e^{i (n \cdot \varphi)}$                                            \\
+            n-te Wurzel         & \hspace*{-10pt} $\sqrt[n]{z} = \sqrt[n]{|z|} \cdot e^{i(\frac{\varphi}{n} + \frac{2 \pi k}{n})}, \, k = 0, \dots, n-1$ \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Fundamentalsatz der Algebra}
+
+    Jedes Polynom $p(z)$ vom Grad $n \geq 1$ hat in $\mathbb{C}$ genau $n$ Nullstellen.
+
+    \begin{center}
+        $p(z) = z^n + a_{n-1}z^{n-1} + \dots + a_1 z + a_0$
+    \end{center}
+
+    Damit ist $\mathbb{C}$ im Unterschied zu $\R$ \textbf{algebraisch vollständig}.
+
+    \subsubsection{Mitternachtsformel}
+
+    Die Nullstellen von $az^2 + b z + c = 0$ sind für $z \in \C$:
+
+    \begin{center}
+        \eqbox{$z_\pm = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-b}{2a} \pm \sqrt{\dfrac{b^2 - 4 ac}{4a^2}}$}
+    \end{center}
+
+    \subsubsection{Binomische Formeln}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.35}
+        \begin{minipage}{0.465 \linewidth}
+            \begin{center}
+                \begin{tabular}{r l}
+                    $(a+b)^2$    & \hspace*{-10pt}$= a^2 + 2ab + b^2$ \\
+                    $(a-b)^2$    & \hspace*{-10pt}$= a^2 - 2ab + b^2$ \\
+                    $(a+b)(a-b)$ & \hspace*{-10pt}$= a^2 - b^2$       \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+        \,\vline\,
+        \begin{minipage}{0.5 \linewidth}
+            \begin{center}
+                \renewcommand{\arraystretch}{1.35}
+                \begin{tabular}{r l}
+                    $(a + b)^3$ & \hspace*{-10pt}$= a^3 + 3a^2 b + 3a b^2 + b^3$ \\
+                    $(a - b)^3$ & \hspace*{-10pt}$= a^3 - 3a^2b + 3 a b^2 - b^3$ \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \subsubsection{Binomischer Lehrsatz}
+
+    Für alle $n \in \N$ und $x,y \in \R$ gilt:
+
+    \begin{center}
+        \eqbox{$\displaystyle (x + y)^n = \sum\limits^n_{k = 0} \binom{n}{k} \, x^k y^{n-k}$} \, wobei $\displaystyle \binom{n}{k} = \dfrac{n!}{k!(n-k)!}$
+    \end{center}
+
+
+    \subsection{Sonstiges}
+
+    Bei Ungleichungen muss man bei einer \textbf{Multiplikation mit negativen Zahlen das Relationszeichen umdrehen}! \medskip
+
+    Zum herausfinden einer Rekursivformel ist eine Primfaktorzerlegung sehr hilfreich!
+
+    \subsubsection{Die Bernouillische Ungleichung}
+
+    $\forall x \in \R, x \geq -1$ und $\forall n \in \N$ gilt:
+
+    \begin{center}
+        $(1 + x)^n \geq 1 + nx$
+    \end{center}
+
+
+    \subsubsection{Wallisches Produkt}
+
+    Die irrationale Zahl $\pi$ wird approximiert durch:
+
+    \begin{center}
+        $\displaystyle \pi = \lim\limits_{n \to \infty} \dfrac{1}{n} \dfrac{(2^n n!)^4}{(2n!)^2}$
+    \end{center}
+
+    \subsubsection{Gerade und Ungerade Funktionen}
+
+    Eine Funktion $f(t)$ heisst:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{l} \toprule
+            i) \emph{gerade} falls $f(t) = f(-t)$ (Symmetrisch zur $y$-Achse).   \\
+            ii) \emph{ungerade} falls $f(t) = -f(-t)$ (Punktsymm. zum Ursprung). \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    Es gelten die folgenden Eigenschaften: \medskip
+
+    - Das Produkt zweier geraden oder ungeraden Funktionen ist gerade.
+
+    - Das Produkt einer geraden und ungeraden Funktion ist ungerade.
+
+    - Falls $f(t)$ gerade ist, gilt $\int\limits_{-a}^{0} f(t) dt = \int\limits_{0}^{a} f(t) dt$.
+
+    - Falls $f(t)$ ungerade ist, gilt $\int\limits_{-a}^{a} f(t) dt = 0$.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Trigonometrische Funktionen: Wertetabelle}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \begin{tabular}{r c c c c c} \toprule
+            deg/rad & 0$^\circ$/0 & 30$^\circ$/$\frac{\pi}{6}$ & $45^\circ$/$\frac{\pi}{4}$ & 60$^\circ$/$\frac{\pi}{3}$ & 90$^\circ$/$\frac{\pi}{2}$ \\ \midrule
+            sin     & $0$         & $\frac{\sqrt{1}}{2}$       & $\frac{\sqrt{2}}{2}$       & $\frac{\sqrt{3}}{2}$       & $1$                        \\
+            cos     & $1$         & $\frac{\sqrt{3}}{2}$       & $\frac{\sqrt{2}}{2}$       & $\frac{\sqrt{1}}{2}$       & $0$                        \\
+            tan     & $0$         & $\frac{\sqrt{3}}{3}$       & $1$                        & $\sqrt{3}$                 & -                          \\ \bottomrule
+        \end{tabular}
+        \begin{tabular}{r c c c c} \toprule
+            deg/rad & 120$^\circ$/$\frac{2\pi}{3}$ & 135$^\circ$/$\frac{3\pi}{4}$ & $150^\circ$/$\frac{5\pi}{6}$ & 180$^\circ$/$\pi$ \\ \midrule
+            sin     & $\frac{\sqrt{3}}{2}$         & $\frac{\sqrt{2}}{2}$         & $\frac{1}{2}$                & $0$               \\
+            cos     & $- \frac{1}{2}$              & $- \frac{\sqrt{2}}{2}$       & $- \frac{\sqrt{3}}{2}$       & $-1$              \\
+            tan     & $-\sqrt{3}$                  & $- 1$         & $-\frac{\sqrt{3}}{3}$        & $0$               \\ \bottomrule
+        \end{tabular}
+    \end{center}
+
+
+    \subsection{Trigonometrische und Hyperbolische Identitäten}
+
+    \begin{center}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \eqboxf{$cos^2(x) + sin^2(x) = 1$} \medskip
+
+                \eqbox{$\tan(x) = \dfrac{\sin(x)}{\cos(x)}$}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \eqbox{$\cosh^2(x) - \sinh^2(x) = 1$} \medskip
+
+                \eqbox{$\tanh(x) = \dfrac{\sinh(x)}{\cosh(x)}$}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \subsubsection{Trigonometrische Additionstheoreme}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{l l} \toprule
+            \multicolumn{2}{l}{$\cos(x\pm y)=\cos(x)\cos(y)\mp\sin(x)\sin(y)$}                                                  \\
+            \multicolumn{2}{l}{$\sin(x\pm y)=\sin(x)\cos(y)\pm\cos(x)\sin(y)$}                                                  \\
+            \midrule
+            $\cos \left(x+\frac{1}{2} \pi\right)=-\sin (x)$  & \hspace*{-10pt} $\sin \left(x+\frac{1}{2} \pi\right)=\cos (x)$   \\
+            \midrule
+            \multicolumn{2}{l}{$\sin(x)\sin(y)=\frac{1}{2}(\cos(x-y)-\cos(x+y))$}                                               \\
+            \multicolumn{2}{l}{$\cos(x)\cos(y)=\frac{1}{2}(\cos(x-y)+\cos(x+y))$}                                               \\
+            \multicolumn{2}{l}{$\sin(x)\cos(y)=\frac{1}{2}(\sin(x-y)+\sin(x+y))$}                                               \\
+                \midrule
+            $\sin^2(x)=\frac{1}{2}(1-\cos(2x))$              & \hspace*{-10pt} $\cos^2(x)=\frac{1}{2}(1+\cos(2x))$              \\
+            $\sin^{3}(x)=\frac{1}{4}(3 \sin (x)-\sin (3 x))$ & \hspace*{-10pt} $\cos^{3}(x)=\frac{1}{4}(3 \cos (x)+\cos (3 x))$ \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \subsubsection{Hyperbolische Additionstheoreme}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{l l} \toprule
+            \multicolumn{2}{l}{$\sinh(x \pm y) = \sinh(x) \cosh(y) \pm \cosh(x) \sinh(y)$} \\
+            \multicolumn{2}{l}{$\cosh(x \pm y) = \cosh(x) \cosh(y) \pm \sinh(x) \sinh(y)$} \\
+            \midrule
+            $\cosh(x) = \cos(ix)$              & $\sinh(x) = -i \sin(ix)$                  \\
+            \midrule
+            $\sinh^2(x)=\frac{\cosh(2x)-1}{2}$ & $\cosh^2(x)=\frac{\cosh(2x)+1}{2}$        \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+    \vfill\null
+    \pagebreak
+
+
+    \section{Folgen und Reihen}
+
+
+    \subsection{Grenzwert einer Folge}
+
+    Die Folge $(a_n)_{n \in \mathbb{N}}$ konvergiert gegen den Grenzwert $a$ für $n \to \infty$, falls
+    \begin{center}
+        \eqboxf{$\forall \epsilon > 0 \,\, \exists N(\epsilon) \in \mathbb{N} \text{ so dass } \forall n \geq N: | a_n - a | < \epsilon$}
+    \end{center}
+
+    Wenn dies gilt, schreibt man: $\liminfty{n} a_n = a$ oder $a_n \to a (n \to \infty)$
+
+    \begin{center}
+        \begin{minipage}{0.5\linewidth}
+            \begin{center}
+                \includegraphics[width=1\linewidth]{Bilder/z_02.jpg}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.47 \linewidth}
+            i) Eine Folge heisst \emph{konvergent}, falls sie einen Grenzwert besitzt. \medskip
+
+            ii) Besitzt die Folge keinen Grenzwert heisst sie \emph{divergent}.
+        \end{minipage}
+    \end{center}
+
+
+    \subsection{Monotonie bei Folgen}
+
+    Eine Folge $(a_n)_{n \in \N}$ bzw. $n \mapsto a_n$ heisst ..., wenn
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \begin{tabular}{r l} \toprule
+            monoton wachsend:        & $a_1 \leq a_2 \leq \dots \leq a_{n-1} \leq a_n$ \\
+            monoton fallend:         & $a_1 \geq a_2 \geq \dots \geq a_{n-1} \geq a_n$ \\
+            streng monoton wachsend: & $a_1 < a_2 < \dots < a_{n-1} < a_n$             \\
+            streng monoton fallend:  & $a_1 > a_2 > \dots > a_{n-1} > a_n$             \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+
+    \subsection{Konvergenzkriterien}
+
+    \subsubsection{Monotone Konvergenz}
+
+    Sei $(a_n)_{n \in \N} \in A$ eine nach oben beschränkte monton wachsende Folge bzw. eine nach unten beschränkte monoton fallende Folge. Dann gilt
+
+    \begin{center}
+        \eqbox{$\liminfty{n} a_n = \begin{cases}
+                    \sup\limits_{n \in \N}(A) & \text{falls monoton wachsend} \\
+                    \inf\limits_{n \in \N}(A) & \text{falls monoton fallend}  \\
+                \end{cases}$}
+    \end{center}
+
+    \subsubsection{Satz: Rechenregeln unter Konvergenzbedingung}
+
+    Seien $(a_n)_{n \in \N}$, $(b_n)_{n \in \N}$ konvergent mit den Grenzwerten $a$ bzw. $b$. Dann
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \begin{tabular}{r l} \toprule
+            i)   & $\liminfty{n}(a_n + b_n) = \liminfty{n} a_n + \liminfty{n} b_n = a + b$                                      \\
+            ii)  & $\liminfty{n}(a_n \cdot b_n) = \liminfty{n} a_n \cdot \liminfty{n} b_n = a \cdot b$                          \\
+            iii) & Falls $\forall n: b \neq 0 \neq b_n$, dann gilt: $\liminfty{n} \left(\dfrac{a_n}{b_n}\right) = \dfrac{a}{b}$ \\
+            iv)  & Falls $a_n \leq b_n$ für alle $n \in \mathbb{N}$, so ist auch $a \leq b$                                     \\ \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \subsubsection{Dominanz}
+
+    Sei $a,b > 1$ und $r \in N \setminus {1}$. Für die \textbf{Stärke der Divergenz} ($\to \infty$) geltet folgende Kette der asymptotischen Dominanz:
+
+    \begin{center}
+        $\lim\limits_{n \to \infty}$ \eqbox{$\log_b(n) \prec \sqrt{n} \prec n \prec n^r \prec a^n \prec n! \prec n^n$}
+    \end{center}
+
+    \subsubsection{Satz}
+
+    Seien $(a_n)_{n \in \N}$ und $(b_n)_{n \in \N}$ zwei Folgen. Sei $\lim\limits_{n \to \infty} a_n = 0$ und $(b_n)_{n \in \N}$ beschränkt. Dann gilt $\lim\limits_{n \to \infty} a_n b_n = 0$.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Teilfolgen und Häufungspunkte}
+
+    \subsubsection{Teilfolge}
+
+    Sei $l: N \to \N$ streng monoton wachsende Abzählung, dann ist $(a_{l(n)})_{n \in \N}$ eine \emph{Teilfolge} von $(a_n)_{n \in \N}$. \medskip
+
+    \subsubsection{Häufungspunkt}
+
+    Ein Punkt $a \in \R$ heisst \emph{Häufungspunkt} von $(a_n)_{n \in \N}$, falls $(a_n)_{n \in \N}$ gegen $a$ eine konvergente Teilfolge besitzt:
+
+    \begin{center}
+        $a = \lim\limits_{l \to \infty} a_{l(n)}$
+    \end{center}
+
+    $a$ ist ein Häufungspunkt von $(a_n)_{n \in \mathbb{N}}$, genau dann wenn
+    \begin{center}
+        \eqbox{$\forall \epsilon > 0 \, \, \forall n \in \mathbb{N} \,\, \exists \, l \geq n: | a - a_{l(n)} | < \epsilon$}
+    \end{center}
+
+    \subsubsection{Limes superior und inferior}
+
+    Sei $(a_n)_{n \in \mathbb{N}}$ eine Folge, dann ist Limes superior und Limes inferior:
+
+    \begin{center}
+        \begin{minipage}{0.5\linewidth}
+            \begin{center}
+                \includegraphics[width=1\linewidth]{Bilder/z_03.jpg}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \renewcommand{\arraystretch}{1.25}
+                \begin{tabular}{r l} \toprule
+                    $\liminf\limits_{n \to \infty} a_n$ & \hspace*{-10pt}$:= \sup\limits_{n \in \mathbb{N}}\inf\limits_{k \geq n} a_k$  \\
+                    $\limsup\limits_{n \to \infty} a_n$ & \hspace*{-10pt}$:= \inf\limits_{n \in \mathbb{N}}\sup\limits_{k \geq n} a_k $ \\
+                    \bottomrule
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    wobei Limes superior und Limes inferior beides Häufungspunkte von $(a_n)_{n \in \N}$ sind. Ausserdem gilt: \medskip
+
+
+    i) Eine beschränkte Folge konvergiert $\Leftrightarrow \limsup\limits_{n \to \infty} a_n = \liminf\limits_{n \to \infty} a_n$
+
+    ii) Eine beschränkte Folge, welche nicht konvergiert, hat mindestens zwei
+    Häufungspunkte.
+
+
+    \subsubsection{Bolzano Weierstrass}
+
+    \begin{center}
+        \eqboxf{\begin{tabular}{C{0.89\linewidth}}
+                Jede beschränkte Folge in $\R^d$ besitzt eine konvergente Teilfolge, also auch einen Häufungspunkt. \\
+            \end{tabular}}
+    \end{center}
+
+
+    \subsection{Cauchy Folge und Cauchy-Kriterium}
+
+    Eine Folge $(a_n)_{n \in \N}$ heisst \emph{Cauchy-Folge}, falls gilt
+
+    \begin{center}
+        \eqbox{$\forall \epsilon > 0 \,\, \exists n_0(\epsilon) \in \N \text{ s.d. } \forall m, n \geq n_0(\epsilon): |a_m - a_n| < \epsilon$}
+    \end{center}
+
+    D.h. wenn es zu jedem $\epsilon > 0$ einen Index $n_0(\epsilon)$ gibt, so dass ab diesem Index alle Folgenglieder weniger als $\epsilon$ voneinander entfernt sind.
+
+    \subsection{Satz: Cauchy-Kriterium}
+
+    Für $(a_n)_{n \in \N} \subset \R$ sind äquivalent:
+
+    \begin{center}
+        \eqboxf{$(a_n)_{n \in \N} \text{ konvergiert } \Leftrightarrow (a_n)_{n \in \N} \text{ ist Cauchy}$}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Folgen in $\R^d$ oder $\mathbb{C}$}
+
+    Sei $(\vect{a}_n)_{n \in \mathbb{N}}$ eine Folge in $\R^d$ mit $\vect{a}_n = (a^1_n, \dots, a^d_n) \in \R^d$. Es gilt
+
+    \begin{center}
+        \eqbox{$\liminfty{n} \vect{a}_n = \vect{a}$ falls $\liminfty{n} || \vect{a}_n - \vect{a} || = 0$}
+    \end{center}
+
+    Es sind äquivalent: $\liminfty{n} \vect{a}_n = \vect{a}  \Leftrightarrow \forall i \in \{1, \dots, d\}:  \liminfty{n} a_n^i = a^i$
+
+    \subsubsection{Beschränkt in $\R^d$}
+
+    Eine  vektorwertige Folge $(\vect{a}_n)_{n \in \mathbb{N}} \subset \R^d$ ist \emph{beschränkt}, falls gilt
+
+    \begin{center}
+        \eqbox{$\exists \, C \in \R \text{ so dass } \forall n \in \mathbb{N}: ||\vect{a}_n|| \leq C$}
+    \end{center}
+
+
+    \subsection{Reihen}
+
+    Sei $(a_k)_{k \in \N}$ eine Folge. Die Folge $(S_n)_{n \in \mathbb{N}}$ der \emph{Partialsummen} ist
+
+    \begin{center}
+        $\displaystyle S_n = a_1 + \dots + a_n = \sum\limits^n_{k = 1} a_k, \quad n \in \N$
+    \end{center}
+
+    Man sagt die Reihe ist \emph{konvergent}, falls $\liminfty{n} S_n = \sum\limits^\infty_{k = 1} a_k$ existiert.
+
+    \subsubsection{Cauchy Kriterium}
+
+    Die Reihe $\sum\limits_{k=1}^\infty a_k$ ist konvergent \emph{genau dann, wenn} gilt:
+
+    \begin{center}
+        \eqbox{$\left| \sum\limits_{k = n+1}^m a_k \right| \to 0 \qquad (n \geq l, l \to \infty)$}
+    \end{center}
+
+
+    \subsection{Konvergenzkriterien für Reihen}
+
+    Die Bedingung Nullfolge ($a_k \xrightarrow[]{k \to \infty} 0$) ist \textbf{notwendig}, aber \emph{nicht} hinreichend für die Konvergenz einer Reihe.
+
+    \subsubsection{Quotientenkriterium}
+
+    Sei $(a_k)_{k \in \N}$ eine Folge in $\R$ oder $\C$. Sei $a_k \neq 0$ und $k \in \N$. Es gilt:
+
+    \begin{center}
+        \eqbox{$\liminfty{k} \left| \dfrac{a_{k+1}}{a_k} \right| = \begin{cases}
+                    < 1 & S_n \text{ konvergiert absolut}, \\
+                    > 1 & S_n \text{ divergiert}           \\
+                \end{cases}$}
+    \end{center}
+
+    \subsubsection{Wurzelkriterium}
+
+    Sei $(a_k)_{k \in \N}$ eine Folge in $\R$ oder $\C$.
+
+    \begin{center}
+        \eqboxf{$\limsup\limits_{k \to \infty} \sqrt[k]{|a_k|} = \begin{cases}
+                    < 1 & S_n \text{ konvergiert absolut}, \\
+                    > 1 & S_n \text{ divergiert}           \\
+                \end{cases}$}
+    \end{center}
+
+    \subsubsection{Minorantenkriterium}
+
+    Sei $b_n \leq a_n$ und $\displaystyle \sum\limits_{n = 1}^\infty b_n$ divergent $\displaystyle \Rightarrow \sum\limits_{n = 1}^\infty a_n$ ist auch divergent.
+
+    \subsubsection{Majorantenkriterium}
+
+    Sei $|a_n| \leq b_n$ und $\displaystyle \sum\limits_{n = 1}^\infty b_n$ konvergent $\displaystyle \Rightarrow \sum\limits_{n = 1}^\infty a_n$ ist auch konvergent.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Absolute Konvergenz}
+
+    Die Reihe $\displaystyle \sum\limits_{n = 1}^\infty a_n$ \emph{konvergiert absolut}, falls $\displaystyle \sum\limits_{n = 1}^\infty | a_n |$ konvergiert.
+
+
+    \subsubsection{Satz}
+
+    Seien die Reihen $\displaystyle \sum\limits_{n = 1}^\infty a_n$ und $\displaystyle \sum\limits_{k = 1}^\infty b_k$ absolut konvergent. Dann konvergiert die Reihe der Produkte absolut mit
+
+    \begin{center}
+        \eqbox{$\displaystyle \sum\limits_{n,k = 1}^\infty a_n b_k = \sum\limits_{n = 1}^\infty a_n \cdot \sum\limits_{k = 1}^\infty b_k$}
+    \end{center}
+
+    unabhängig von der Summationsreihenfolge.
+
+
+    \subsection{Satz: Leibnitzkriterium}
+
+    Sei $(a_n)_{n \in \N}$ eine monoton fallende, reelle \emph{Nullfolge}. Dass heisst
+
+    \begin{center}
+        \eqbox{$a_{n + 1} \leq a_n \quad \forall n \in \N_0$ und $\lim\limits_{n \to \infty} a_n = 0$}
+    \end{center}
+
+    Dann ist die Reihe $\displaystyle \sum\limits_{n = 0}^\infty (-1)^n a_n$ konvergent.
+
+
+    \subsection{Standard Reihenabschätzung}
+
+    \begin{center}
+        \eqbox{$\displaystyle \left| \sum\limits_{k = 1}^n a_k \right| \leq n \cdot \max\limits_{1 \leq k \leq n} | a_k |$}
+    \end{center}
+
+
+    \subsection{Geometrische Reihe}
+
+    Die Geometrische Reihe ist für $|z| < 1$ konvergent und es gilt:
+
+    \begin{center}
+        \eqboxf{$\displaystyle\sum\limits_{k = 0}^\infty z^k = \dfrac{1}{1-z}$} \qquad\qquad \eqbox{$\displaystyle\sum\limits_{k = 0}^n z^k = \dfrac{1 - z^{n + 1}}{1 - z}$}
+    \end{center}
+
+
+    \subsection{Wichtige Reihen}
+
+    \begin{tabular}{r c l} \toprule
+        Harmonische Reihe:           & \hspace*{-10pt} $\displaystyle\sum\limits^\infty_{k = 1} \dfrac{1}{k}$             & divergent                                                                               \\
+        Riemann'sche $\zeta$-Funkt.: & \hspace*{-10pt} $\displaystyle\zeta(s) =\sum\limits^\infty_{k = 1} \dfrac{1}{k^s}$ & $\begin{cases} 0 < s \leq 1 & \text{divergent} \\ 1 < s & \text{konvergent}\end{cases}$ \\
+        \bottomrule
+    \end{tabular}
+
+
+    \subsection{Wichtige Potenzreihen}
+
+    Folgende Funktionen besitzen für alle $z \in \C$ konvergente Potenzreihen:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \begin{tabular}{r l}
+                    $\exp(z)$ & \hspace*{-10pt}$ := \displaystyle\sum\limits_{n = 0}^\infty \dfrac{z^n}{n!}$                 \\
+                    $\sin(z)$ & \hspace*{-10pt}$:= \displaystyle\sum\limits_{n = 0}^\infty (-1)^n \dfrac{z^{2n+1}}{(2n+1)!}$ \\
+                    $\cos(z)$ & \hspace*{-10pt}$:= \displaystyle\sum\limits_{n = 0}^\infty (-1)^n \dfrac{z^{2n}}{(2n)!}$     \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \begin{tabular}{r l}
+                    $\sinh(z)$ & \hspace*{-10pt}$:= \displaystyle\sum\limits_{n = 0}^\infty \dfrac{z^{2n+1}}{(2n+1)!}$ \\
+                    $\cosh(z)$ & \hspace*{-10pt}$:= \displaystyle\sum\limits_{n = 0}^\infty \dfrac{z^{2n}}{(2n)!}$     \\
+                \end{tabular}
+            \end{center}
+        \end{minipage}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Potenzreihen}
+
+    Sei $z \in \C$. Eine Reihe der folgenden Form nennt man eine Potenzreihe:
+
+    \begin{center}
+        \eqbox{$\displaystyle p(z) := c_0 + c_1 z + c_2 z^2 + \dots = \sum\limits_{k = 0}^\infty c_k z^k$}
+    \end{center}
+
+    \subsubsection{Konvergenzradius}
+
+    Eine Potenzreihe ist konvergent für alle $|z| < \rho$ und es gilt:
+
+    \begin{center}
+        \eqboxf{$\rho := \dfrac{1}{\limsup\limits_{k \to \infty} \sqrt[k]{|c_k|}}$} $\begin{cases}
+                |z| < \rho & \text{konvergiert absolut} \\
+                |z| = \rho & \text{keine Aussage}       \\
+                |z| > \rho & \text{divergiert}          \\
+            \end{cases}$
+    \end{center}
+
+    Innerhalb von $\rho$ darf man Limes, Ableitung, Integral austauschen!
+
+
+    \subsubsection{Potenzreihen konvergieren gleichmässig}
+
+    Sei eine Potenzreihe $p(z)$ mit Konvergenzradius $\rho > 0$. Dann konvergiert
+
+    \begin{center}
+        \eqbox{$\displaystyle p_n(z) = \sum\limits_{k = 0}^{n - 1} a_k z^k$}
+    \end{center}
+
+    gleichmässig gegen $p(z)$ auf $B_r(0)$ für jedes $r < p$.
+
+
+    \subsubsection{Potenzreihen sind stetig}
+
+    Potenzreihen sind \emph{stetig} im Inneren ihres Konvergenzradius $\rho$.
+
+
+    \subsubsection{Potenzreihen sind differenzierbar}
+
+    Eine Potenzreihe $f(x) = \sum\limits_{k = 0}^{\infty} a_k x^k$ ist im Inneren ihres Konvergenzradius \emph{gliedweise} differenzierbar. Die Ableitung von $f(x)$ ist
+
+    \begin{center}
+        \eqbox{$\displaystyle f'(x) = \sum\limits_{k = 1}^{\infty} k a_k x^{k - 1}$}
+    \end{center}
+
+    Ausserdem bestizt die Ableitung $f'(x)$ den \emph{gleichen} Konvergenzradius. \medskip
+
+    \textbf{Achtung}: Oft ist es sinnvoll die Ableitungen der einzelnen Potenzen kurz anzuschauen, damit man die Formel nicht falsch anwendet!
+
+
+    \subsubsection{Potenzreihen sind integrierbar}
+
+    Eine Potenzreihen $f(x) = \sum\limits_{k = 0}^{\infty} a_k x^k$ ist innerhalb ihres Konvergenzradius \emph{gliedweise} integrierbar. Das Integral von $f(x)$ ist
+
+    \begin{center}
+        \eqbox{$\displaystyle \int\limits f(x) dx = \sum\limits_{k = 0}^{\infty} \dfrac{a_k}{k + 1} x^{k + 1}$}
+    \end{center}
+
+    und die Stammfunktion $F(x)$ besitzt den \emph{gleichen} Konvergenzradius $\rho$. \medskip
+
+    \textbf{Achtung}: Oft ist es sinnvoll das Integral mit den einzelnen Potenzen kurz anzuschauen, damit man die Formel nicht falsch anwendet!
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Wichtige Grenzwerte}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{2}
+        \begin{tabular}{l l l l l} \toprule
+            $\lim\limits_{n \to \infty} (1 + \frac{1}{n})^n$                  & \hspace*{-10pt} $= e$      & \hspace*{+10pt}
+            $\lim\limits_{n \to \infty} (1 + \frac{a}{n})^n$                  & \hspace*{-10pt}$= e^a$                       \\
+            $\lim\limits_{n \to \infty} n ( a^{\frac{1}{n}} - 1)$             & \hspace*{-10pt}$= \log(a)$ & \hspace*{+10pt}
+            $\lim\limits_{n \to 0} \frac{a^n - 1}{n}$                         & \hspace*{-10pt}$= \log(a)$                   \\
+            $\lim\limits_{n \to \infty} \sqrt[n]{n}$                          & \hspace*{-10pt}$= 1$       & \hspace*{+10pt} \\
+            \midrule
+            $\lim\limits_{t \to \infty} t \sin(\frac{1}{t})$                  & \hspace*{-10pt}$= 1$       & \hspace*{+10pt}
+            $\lim\limits_{t \to 0} \frac{\sin(t)}{t}$                         & \hspace*{-10pt}$= 1$                         \\
+            $\lim\limits_{t \to \infty} t \log(1 + \frac{1}{t})$              & \hspace*{-10pt}$= 1$       & \hspace*{+10pt}
+            $\lim\limits_{t \to 0} \frac{\log(1 + t)}{t}$                     & \hspace*{-10pt}$= 1$                         \\
+            $\lim\limits_{t \to \infty} \frac{1}{t^2(1 - \cos(\frac{1}{t}))}$ & \hspace*{-10pt}$= 1$       & \hspace*{+10pt}
+            $\lim\limits_{t \to 0} \frac{t^2}{1 - \cos(t)}$                   & \hspace*{-10pt}$= 2$                         \\
+            $\lim\limits_{t \to \infty} \frac{1}{t \sin(\frac{1}{t})}$        & \hspace*{-10pt}$= 1$       & \hspace*{+10pt}
+            $\lim\limits_{t \to 0} \frac{t}{\sin(t)}$                         & \hspace*{-10pt}$= 1$                         \\
+            $\lim\limits_{t \to 0} \frac{e^t - 1}{t}$                         & \hspace*{-10pt}$= 1$       & \hspace*{+10pt} 
+            $\lim\limits_{t \to 0} \frac{\log(1 + 2t)}{\log(1 + t)}$ & \hspace*{-10pt}$= 2$ \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+
+    \subsection{Tipps Grenzwertberechnung}
+
+    Verschiedene mögliche Ansätze:
+
+    \begin{itemize}
+        \item Bei Grenzwerten, welche eingesetzt $\frac{0}{0}$ oder $\frac{\infty}{\infty}$ geben, L'hopital anwenden!
+        \item Wurzelterme: 3te Binomische Formel versuchen
+        \item Schwieriger $\lim\limits_{n \to 0} (\dots)$ : Taylorformel mit Entwicklungspunkt $0$ benutzen (getrennt für Nenner und Zähler anwenden!!!). Dies funktioniert, da die Approximation im Entwicklungspunkt exakt ist.
+        \item Grenzwerten mit vielen Funktionen: So umformen zu versuchen, dass man die Grenzwerte unter 'Wichtige Grenzwerte' verwenden kann!
+              \begin{itemize}
+                  \item Bei $\lim\limits_{n \to \infty} (\dots)^n$ muss man fast immer ausschliesslich $e^{n \cdot log(\dots)}$ als erste Umformung benutzen!
+              \end{itemize}
+    \end{itemize}
+
+    \subsubsection{Grenzwert und Kompositionen stetiger Funktionen}
+
+    Wird der Grenzwert einer Komposition stetiger Funktionen genommen, so darf man den Grenzwert auf die innere Funktion anwenden. Beispiel:
+
+    \begin{center}
+        $\lim\limits_{x \to 0} \exp(\frac{1}{x}\log(\cos(x))) = \exp(\lim\limits_{x \to 0} \frac{1}{x}\log(\cos(x)) )$
+    \end{center}
+    \vfill\null
+    \pagebreak
+
+
+
+    \section{Stetigkeit auf $\R$ und $\R^d$}
+
+
+    \subsection{Grenzwert einer Funktion}
+
+    \subsubsection{Der Abschluss}
+
+    Sei $\Omega \in \R^d$. Der Abschluss von $\Omega$ ist die Menge:
+
+    \begin{center}
+        \eqbox{$\overline{\Omega} = \{x \in \R^d; \exists (x_k)_{k \in \N} \subset \Omega, \, \lim\limits_{k \to \infty} x_k = x \}$}
+    \end{center}
+
+    In Worten formuliert: Der Abschluss sind alle Punkte $x_0$, die durch Punkte in $\Omega$ erreichbar sind. \medskip
+
+    Bem: Offenbar gilt $\Omega \subset \overline{\Omega}$.
+
+    \subsubsection{Definition: Grenzwert einer Funktion}
+
+    Sei $\Omega \subset \R^d$, $f: \Omega \to \R^n$ und $x_0 \in \overline{\Omega}$. $f$ hat an der Stelle $x_0$ den \emph{Grenzwert} $a \in \R^n$, falls für jede Folge $(x_k)_{k \in \N}$ in $\Omega$ mit
+
+    \begin{center}
+        $x_k \xrightarrow[]{k \to \infty} x_0$ gilt $f(x_k) \xrightarrow[]{k \to \infty} a$.
+    \end{center}
+
+    Ist dies der Fall und $x_0 \in \Omega$, dann muss gelten $\lim\limits_{x \to x_0} f(x) = f(x_0)$.
+
+    \subsubsection{Stetig in $x_0$ und stetig ergänzbar}
+
+    Sei $\Omega \subset \R^d$, $f: \Omega \to \R^n$. Man sagt: \medskip
+
+    i) Sei $x_0 \in \Omega$. $f$ ist \emph{stetig} in $x_0$, falls $f$ in $x_0$ einen Grenzwert besitzt. \medskip
+
+    ii) Sei $x_0 \in \overline{\Omega} \setminus \Omega$. $f$ heisst an der Stelle $x_0$ \emph{stetig ergänzbar}, falls $f$ in $x_0$ einen Grenzwert besitzt. Notation: $\lim\limits_{k \to \infty} f(x_k) = a \, \Leftrightarrow \, \lim\limits_{\substack{x \to x_0 \\ x \neq x_0}} = a$
+
+
+    \subsection{Für $\R$: Links- und Rechtsseitiger Grenzwert}
+
+    \begin{center}
+        \begin{minipage}{0.48\linewidth}
+            Nähert man sich von Links an $x_0$ an, d.h. $x < x_0$ dann gilt:
+
+            \begin{center}
+                \eqbox{$f(x_0^-) := \lim\limits_{x \to x_0^-} f(x)$}
+            \end{center}
+        \end{minipage}\,\,
+        \begin{minipage}{0.48\linewidth}
+            Nähert man sich von Rechts an $x_0$ an, d.h. $x > x_0$ dann gilt:
+
+            \begin{center}
+                \eqbox{$f(x_0^+) := \lim\limits_{x \to x_0^+} f(x)$}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    $f$ ist stetig an der Stelle $x_0$ genau dann, wenn $f(x_0^-) = f(x_0^+) = f(x_0)$.
+
+    \subsubsection{Satz}
+
+    Sei $]a,b[ \to \R$ monoton wachsend oder monoton fallend. Dann existieren für jedes $x_0 \in ]a,b[$ die links- und rechtsseitigen Grenzwerte.
+
+
+    \subsection{Für $\R$: Monotonie bei Funktionen}
+
+    $f:[a,b] \to \R$ heisst streng monoton wachsend, falls gilt
+
+    \begin{center}
+        \eqbox{$a \leq x < y \leq b \Rightarrow f(x) < f(y)$}
+    \end{center}
+
+    $f:[a,b] \to \R$ heisst streng monoton fallend, falls gilt
+
+    \begin{center}
+        \eqbox{$a \leq x < y \leq b \Rightarrow f(x) > f(y)$}
+    \end{center}
+
+
+    \subsection{Für $\R^d$: Grenzwert in $\R^d$}
+
+    Sei $(x_k)_{k \in \N} \in \R^d$. Die Folge $x_k$ konvergiert gegen $x$, falls
+
+    \begin{center}
+        \eqbox{$\lim\limits_{k \to \infty} ||x_k - x|| = 0 \, \Leftrightarrow \, \lim\limits_{k \to \infty} x_k^j = x^j$}
+    \end{center}
+
+    wobei $x_k = (x^1_k, \dots, x^d_k)$ und $x = (x^1, \dots, x^d)$ beide in $\R^d$ definiert sind.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Stetige Funktionen}
+
+    Sei $\Omega \subset \R^d$, $f: \Omega \to \R^n$. Dann sagt man:
+
+    \begin{center}
+        $f$ heisst \textbf{stetig auf} $\Omega \subset \R$, falls $f$ in jedem Punkt $x_0 \in \Omega$ stetig ist.
+    \end{center}
+
+    \subsubsection{Satz: Vektorraum $C^0(\Omega, \R)$}
+
+    Sei $\alpha, \beta \in \R$ und $f,g: \Omega \subset \R^d \to \R^n$ stetig. Dann ist $\alpha f + \beta g$ auch stetig. Die stetigen Funktionen $f: \Omega \to \R^n$ bilden also einen $\R$-Vektorraum.
+
+    \begin{center}
+        Notation: \eqbox{$C^0(\Omega, \R) = \{ f: \Omega \to \R^n; f \text{ ist stetig} \}$}
+    \end{center}
+
+    \subsubsection{Satz}
+
+    Sind $f,g \in C^0(\Omega \subset \R^d, \R^n)$. So ist auch $f \circ g \in C^0(\Omega \subset \R^d, \R^n)$.
+
+
+    \subsection{Lipschitz stetig}
+
+    Sei $f: \Omega \subset \R^d \to \R^n$. $f$ heisst \emph{L-Lipschitz stetig} mit der Lipschitzkonstante $0 \leq L$, falls
+
+    \begin{center}
+        \eqbox{$\forall \vect{x},\vect{y} \in \Omega$ gilt $||f(\vect{x}) - f(\vect{y}) || \leq L \cdot || \vect{x} - \vect{y} ||$}
+    \end{center}
+
+    Bemerkung: Lipschitz stetig $\Rightarrow$ Gleichmässig stetig $\Rightarrow$ $f$ ist stetig
+
+
+    \subsubsection{Satz}
+
+    $f: \Omega \subset \R^d \to \R^n$ \emph{L-Lipschitz stetig} $\Rightarrow$ $f$ ist stetig ergänzbar in $x_0 \in \overline{\Omega}$.
+
+
+    \subsection{Kompakt}
+
+    $K \subset \R^d$ heisst kompakt, falls jede Folge $(x_k)_{k \in \N} \subset K$ einen Häufungs-punkt in $K$ besitzt. Ausserdem gilt folgende Äquivalenz:
+
+    \begin{center}
+        \eqbox{K kompakt $\Leftrightarrow$ $K$ ist beschränkt und abgeschlossen.}
+    \end{center}
+
+    Bem: Das eine Menge nicht Kompakt ist, zeigt man am besten, indem man eine unbeschränkte Folge findet (Folge ohne HP).
+
+
+    \subsubsection{Lemma}
+
+    Sei $K \subset \R$ kompakt. Dann ist $K$ beschränkt und es $\exists a,b \in K$ mit
+
+    \begin{center}
+        \eqbox{$-\infty < a = \inf(K) = \min(K) \qquad\quad \max(K) = \sup(K) = b < \infty$}
+    \end{center}
+
+
+    \subsubsection{Satz: Extremumsatz}
+
+    Sei $K \subset \R^d$ kompakt, $f: K \to \R^n$ stetig. Dann ist auch das Bild der Funktion $f: K \to \R^n$ kompakt. Insbesondere gilt:
+
+    \begin{center}
+        \eqboxf{$f: K \to \R^n$ nimmt ihr \emph{Maximum und Minimum} auf $K$ an}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Für $\R$: Weierstrass'sches Kriterium für Stetigkeit}
+    %jelkeismygirlfriend and i love her very much, she a cutie
+
+    $f: \Omega \in \R$ ist an der Stelle $x_0$ stetig \emph{genau dann, wenn}
+
+    \begin{center}
+        \eqboxf{$\forall \epsilon > 0 \,\, \exists \delta > 0 \text{ s.d. } \forall x \in \Omega: \, |x - x_0| < \delta \Rightarrow ||f(x) - f(x_0) || < \epsilon$}
+    \end{center}
+
+    \begin{center}
+        \includegraphics[width=0.75\linewidth]{Bilder/Delta_Epsilon_Kriterium.png}
+    \end{center}
+
+
+    \subsection{Für $\R$: Der Zwischenwertsatz}
+
+    Sei $a < b$, $f:[a,b] \to \R$ stetig. Dann gilt
+
+    \begin{center}
+        \eqboxf{$\forall y \in [f(a),f(b)] \,\, \exists x \in [a,b]$ mit $f(x) = y$}
+    \end{center}
+
+    In Worten: ''Das Bild einer stetigen Funktion, die auf einem Intervall definiert ist, ist ein Intervall.''
+
+    \subsubsection{Satz}
+
+    Sei $f: [a,b] \to \R$ stetig und streng monoton wachsend/fallend. Setze $f(a) = c$ und $f(b) = d$. Dann gilt
+
+    \begin{center}
+        $f:[a,b] \to [c,d]$ ist bijektiv, und $f^{-1}$ ist stetig sowie streng monoton wachsend/fallend
+    \end{center}
+
+
+    \subsubsection{Satz}
+
+    Sei $f:]a,b[ \to \R$ stetig und streng monoton wachsend/fallend mit
+
+    \begin{center}
+        $-\infty \leq c:= \lim\limits_{x \to a^+} f(x) < \lim\limits_{x \to b^-} f(x) =: d \leq \infty$
+    \end{center}
+
+    Dann ist $f:]a,b[ \to ]c,d[$ bijektiv, und $f^{-1}$ ist stetig sowie streng monoton wachsend/fallend.
+
+
+    \subsection{Gleichmässige Stetigkeit}
+
+    Sei $\Omega \subset \R^d$. $f: \Omega \to \R^n$ heisst gleichmässig stetig, falls gilt:
+
+    \begin{center}
+        \eqbox{$\forall \epsilon > 0 \,\, \exists \delta > 0 \text{ s.d. } \forall x,y \in \Omega: \, |x - y| \leq \delta \Rightarrow || f(x) - f(y) || < \epsilon$}
+    \end{center}
+
+    Bemerkung: Lipschitz stetig $\Rightarrow$ Gleichmässig stetig $\Rightarrow$ $f$ ist stetig
+
+    \subsubsection{Satz}
+
+    Sei $f: \Omega \subset \R^d \to \R^n$ gleichmässig stetig $\Rightarrow$ $f$ ist beschränkt auf $\Omega$.
+
+    \subsubsection{Sätze}
+
+    i) Sei $K \subset \R^d$ kompakt und $f \in C^0(K, \R^n) \Rightarrow f$ ist gleichmässig stetig. \medskip
+
+    ii) Sei $f: \Omega \to \R^n$ gleichmässig stetig $\Rightarrow$ $f$ ist auf $\overline{\Omega}$ stetig ergänzbar. \medskip
+
+    iii) Sei $\Omega \subset \R^d$ beschränkt, $f: \Omega \to \R^n$ stetig und auf $\overline{\Omega}$ stetig ergänzbar. $\Rightarrow$ $f$ ist gleichmässig stetig.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Punktweise und gleichmässige Konvergenz}
+
+
+    \subsubsection{Supremumsnorm}
+
+    Sei $f \in C^0(\Omega, \R^n)$. Dann ist die Supremumsnorm
+
+    \begin{center}
+        \eqbox{$||f||_{C^0} := \sup\limits_{x \in \Omega}|| f(x)|| < \infty$}
+    \end{center}
+
+
+    \subsubsection{Punktweise Konvergenz}
+
+    Sei $\Omega \subset \R^d$, und $f, f_k: \Omega \to \R^n,\, k \in \N$. Die Folge $(f_k)_{k \in \N}$ konvergiert punktweise gegen $f$, falls
+
+    \begin{center}
+        \eqbox{$\forall x \in \Omega: \,\lim\limits_{k \to \infty} f_k(x) = f(x)$}
+    \end{center}
+
+
+    \subsubsection{Gleichmässige Konvergenz}
+
+    Sei $\Omega \subset \R^d$, und $f, f_k: \Omega \to \R^n,\, k \in \N$. Die Folge $(f_k)_{k \in \N}$ konvergiert gleichmässig gegen $f$, falls
+
+    \begin{center}
+        \eqboxf{$\lim\limits_{k \to \infty} \sup\limits_{x \in \Omega} || f_k(x) - f(x) || = 0$}
+    \end{center}
+
+    Bemerkung: Gleichmässige Konvergenz $\Rightarrow$ Punktweise Konvergenz
+
+    \subsubsection{Satz}
+
+    Sei $f_k \in C^0(\Omega, \R^n)$ und $f_k$ konvergiert gleichmässig $\Rightarrow$ $f$ ist auf $\Omega$ stetig.
+
+
+    \subsection{Einschub: De Morgansche Regeln}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \eqbox{$(A_1 \cup A_2)^C = A_1^c \cap A_2^c$}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.47 \linewidth}
+            \begin{center}
+                \eqbox{$(A_1 \cap A_2)^C = A_1^c \cup A_2^c$}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+
+    \vfill\null
+    \columnbreak
+
+
+
+    \section{Topologie}
+
+
+    \subsection{Offene Mengen}
+
+    \subsubsection{Der offene Ball}
+
+    Sei $x_0 \in \R^d$ mit $r > 0$. Der offene Ball mit Radius $r$ und Zentrum $x_0$ ist
+
+    \begin{center}
+        \eqbox{$B_r(x_0) := \{x \in \R^d; ||x - x_0|| < r\}$}
+    \end{center}
+
+    \subsubsection{Definition: Offene Menge und Innerer Punkt}
+
+    Ein Punkt $x_0 \in \Omega$ heisst \emph{innerer Punkt} von der Menge $\Omega$, falls
+
+    \begin{center}
+        \eqboxf{$\exists r > 0: B_r(x_0) \subset \Omega$}
+    \end{center}
+
+    Die Menge $\Omega \subset \R^d$ heisst \textbf{offen}, falls jedes $x_0 \in \Omega$ ein innerer Punkt ist.
+
+
+    \subsubsection{Satz: Eigenschaften offener Mengen}
+
+    Es gelten folgenden Eigenschaften für offene Mengen: \medskip
+
+    i) $\Omega_1, \Omega_2 \subset \R^d$ offen $\Rightarrow$ Schnittmenge $\Omega_1 \cap \Omega_2$ ist offen. \medskip
+
+    ii) $\Omega_l \subset \R^d$ offen, $l \in I$ $\Rightarrow$ $\bigcup\limits_{i \in I} \Omega_l$ offen.
+
+    iii) Endliche offene Mengen $(A_i)_{i \in I} \Rightarrow \bigcap\limits_{i \in I} A_i$ offen.
+
+
+    \subsection{Abgechlossene Mengen}
+
+    Eine Menge $A \subset \R^d$ heisst \textbf{abgeschlossen}, falls das Komplement $A^C = \R^d \setminus A$ offen ist.
+
+    \subsubsection{Satz: Eigenschaften abgeschlossener Mengen}
+
+    i) $A_1, A_2$ abgeschlossen $\Rightarrow$ Vereinigungsmenge $A_1 \cup A_2$ abgeschlossen. \medskip
+
+    ii) $A_l \subset \R^d$ abgeschlossen, $l \in I$ $\Rightarrow$ $\bigcap\limits_{i \in I} A_l$ abgeschlossen. \medskip
+
+    iii) Endliche abgeschlossene Mengen $(A_i)_{i \in I} \Rightarrow \bigcup\limits_{i \in I} A_i$ abgeschlossen.
+
+
+    \subsubsection{Bemerkungen}
+
+    i) Die zwei Mengen $\R^n, \, \emptyset$ sind sowohl offen, als auch abgeschlossen. \medskip
+
+    ii) Es gibt Mengen, die weder offen noch abgeschlossen sind!
+
+
+    \subsection{Das Innere, der Abschluss und der Rand einer Menge}
+
+
+    \subsubsection{Das Innere}
+
+    Die Menge der inneren Punkte von $\Omega$
+
+    \begin{center}
+        \eqbox{$int(\Omega) = \overset{\circ}{\Omega} := \{x \in \Omega; \exists r > 0$ s.d. $B_R(x) \subset \Omega \} = \bigcup\limits_{U \subset \Omega, \text{ U offen}} U$}
+    \end{center}
+
+    heisst \textbf{offener Kern} oder das \textbf{Innere} von $\Omega$.
+
+
+    \subsubsection{Der Abschluss}
+
+    Der \textbf{Abschluss} einer Menge ist
+
+    \begin{center}
+        \eqbox{$clos(\Omega) = \overline{\Omega} := \bigcap\limits_{A \subset \Omega, \text{ A abgeschlossen}} A$} \medskip
+
+        $\Leftrightarrow$ \eqbox{$\overline{\Omega} = \{x_0 \in \R^d; \, \exists (x_k)_{k \in \N} \subset \Omega, \, \lim\limits_{k \to \infty} x_k = x_0 \}$}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsubsection{Der Rand}
+
+    Der \textbf{Rand} einer Menge ist
+
+    \begin{center}
+        \eqbox{$\partial\Omega := \overline{\Omega} \setminus \overset{\circ}{\Omega} = \{x \in \R^d; \forall r > 0: B_r(x) \,\cap\, \Omega \neq \emptyset \neq B_r(x) \setminus \Omega\}$}
+    \end{center}
+
+
+    \subsubsection{Satz: Eigenschaften}
+
+    i) Der Rand $\partial \Omega$ ist abgeschlossen.
+
+    ii) Aus $\overset{\circ}{\Omega} \subseteq \Omega \subseteq \overline{\Omega}$ folgt $\overline{\Omega} = \overset{\circ}{\Omega} \cup \partial \Omega$ und die Zerlegung ist disjunkt. \medskip
+
+    iii) Es folgt das Kriterium: \eqboxf{$\Omega$ abgeschlossen $\Leftrightarrow \Omega = \overline{\Omega} \Leftrightarrow \partial \Omega \subset \Omega$}
+
+    iv) Es gilt ausserdem $\overline{\overline{\Omega}} = \overline{\Omega}$, sowie $\overset{\circ}{\overset{\circ}{\Omega}} = \overset{\circ}{\Omega}$
+
+
+    \subsection{Topologisches Kriterium für Stetigkeit}
+
+    Sei $\Omega \subset \R^d$, $x_0 \in \Omega$. Man sagt:
+
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{minipage}{0.7\linewidth}
+            i) $U \subset \Omega$ heisst \textbf{Umgebung} von $x_0$ relativ zu $\Omega$, falls
+
+            \begin{center}
+                \eqbox{$\exists r > 0 \,\text{ mit }\, (B_r(x_0) \cap \Omega) \subset U$}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.27\linewidth}
+            \begin{center}
+                \includegraphics[width = 1\linewidth]{Bilder/Umgebung.jpg}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    ii) Sei $E \subset \R^d$ offen. $U \subset \Omega$ heisst \textbf{relativ offen}, falls $U = E \cap \Omega$. \medskip
+
+    iii) $A \subset \Omega$ heisst \textbf{relativ abgeschlossen}, falls $\Omega \setminus A$ relativ offen ist.
+
+
+
+
+    \subsubsection{Satz: Weierstrass'sches Kriterium für Stetigkeit}
+    % https://de.wikipedia.org/wiki/Stetige_Funktion#Stetigkeit_in_der_Topologie
+
+    Sei $f: \Omega \subset \R^d \to \R^n$ und $x_0 \in \Omega$. Es sind äquivalent:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \begin{tabular}{r p{0.85\linewidth}} \toprule
+                              & \hspace*{-10pt} $f$ ist an der Stelle $x_0$ stetig gemäss Folgekriterium.                                                                                              \\
+            $\Leftrightarrow$ & \hspace*{-10pt} $\forall \epsilon > 0 \,\, \exists \delta > 0 \,\text{ s.d }\, \forall x \in \Omega: \, ||x - x_0|| < \delta \Rightarrow ||f(x) - f(x_0)|| < \epsilon$ \\
+            $\Leftrightarrow$ & \hspace*{-10pt} Für jede Umgebung $V$ von $f(x_0)$ in $\R^n$ ist $U = f^{-1}(V)$ eine Umgebung von $x_0$ in $\Omega$.                                                  \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+
+    \subsubsection{Topologisches Kriterium für Stetigkeit}
+
+    Für $f: \Omega \to \R^n$ sind äquivalent:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \eqboxf{\begin{tabular}{r p{0.8\linewidth}}
+                                  & \hspace*{-10pt} $f$ ist stetig ($f \in C^0$).                                                                      \\
+
+                $\Leftrightarrow$ & \hspace*{-10pt} Das Urbild $U = f^{-1}(V)$ jeder offenen Menge $V \subset \R^n$ ist relativ offen.                 \\
+                $\Leftrightarrow$ & \hspace*{-10pt} Das Urbild $A = f^{-1}(B)$ jeder abgeschlossene Menge $B \subset \R^n$ ist  relativ abgeschlossen. \\
+            \end{tabular}}
+    \end{center}
+
+    \begin{center}
+        \includegraphics[width = 1\linewidth]{Bilder/TopoStetigkeit.png}
+    \end{center}
+    \vfill\null
+    \pagebreak
+
+
+
+    \section{Differentialrechnung auf $\R$}
+
+    \subsection{Differential}
+
+    Sei $f: \Omega \to \R$ mit $\Omega \subset \R$ offen und $x_0 \in \Omega$. $f$ heisst \emph{differenzierbar} an der Stelle $x_0$ falls folgender Grenzwert exisitert:
+
+    \begin{center}
+        \eqbox{$\dfrac{d f}{d x}(x_0) = f'(x_0) := \displaystyle \lim\limits_{h \to 0} \dfrac{f(x_0 + h) - f(x_0)}{h}$}
+    \end{center}
+
+    $f'(x)$ heisst \emph{Ableitung} (oder Differential) von $f$ an der Stelle $x_0$. \medskip
+
+    Bemerkung: Analoges gilt für $f: \Omega \to \R^n$ (Vektorwertige Funktion), einfach Komponentenweise.
+
+    \subsubsection{Geometrische Bedeutung}
+
+    i) Der Differentialquotient $\frac{f(x) - f(x_0)}{x - x_0}$ entspricht der Steigung der Sekante durch die Punkte $(x,f(x)), (x_0, f(x_0))$. \medskip
+
+    ii) Die Ableitung $f'(x_0)$ entspricht der Steigung der Tangente im Punkt $(x_0,f(x_0))$.
+
+    \subsubsection{Satz}
+
+    $f: \Omega \to \R$ differenzierbar an der Stelle $x_0 \in \Omega$ $\Rightarrow$ $f$ ist stetig in $x_0$.
+
+    \subsubsection{Eigenschaften des Differentials}
+
+    Seien $f,g: \Omega \to \R$ an der Stelle $x_0 \in \Omega$ differenzierbar. Dann gilt
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.75}
+        \begin{tabular}{r l} \toprule
+            Summenregel   & $(f(x_0) + g(x_0))' = f'(x_0) + g'(x_0)$                                                 \\
+            Produktregel  & $(f(x_0) \cdot g(x_0))' = f'(x_0) g(x_0) + f(x_0)g'(x_0)$                                \\
+            Quotientregel & $\left(\frac{f(x_0)}{g(x_0)}\right)' = \dfrac{f'(x_0) g(x_0) - g'(x_0)f(x_0)}{g^2(x_0)}$ \\
+            Kettenregel   & $(g(x_0) \circ f(x_0))' = g'(f(x_0)) \cdot f'(x_0)$                                      \\ \bottomrule
+        \end{tabular}
+    \end{center}
+
+    Bei der Quotientregel muss natürlich $g(x_0) \neq 0$ sein.
+
+
+    \subsection{Der Mittelwertsatz}
+
+    Seien $-\infty < a < b < \infty$. Sei $f:[a,b] \to \R$ stetig und differenzierbar in $]a,b[$. Dann existiert $x_0 \in ]a,b[$ mit
+
+    \begin{center}
+        $f(b) = f(a) + f'(x_0)(b - a)$ $\quad \Leftrightarrow \quad$ \eqboxf{$f'(x_0) = \dfrac{f(b) - f(a)}{b - a}$}
+    \end{center}
+
+    \subsubsection{Korollar}
+
+    Sei $f:[a,b] \to \R$ stetig und differenzierbar auf $]a,b[$. Dann gilt:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \begin{tabular}{r l} \toprule
+            i)   & Falls $f'(x) \equiv 0$ auf $]a,b[$, dann ist $f$ konstant           \\
+            ii)  & Falls $f'(x) \geq 0$ auf $]a,b[$, dann ist $f$ monoton wachsend.    \\
+            iii) & Falls $f'(x) > 0$ auf $]a,b[$, dann ist $f$ streng monton wachsend. \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Bernouilli de l'Hôpital}
+
+    Seien $f,g: [a,b] \to \R$ stetig und differenzierbar in $]a,b[$. Sei $g'(x_0) \neq 0$ für alle $x \in ]a,b[$ und $f(a) = 0 = g(a)$ oder $f(a) = \pm \infty = g(a)$. Existiert
+
+    \begin{center}
+        $\lim\limits_{x \to a^+} \dfrac{f'(x)}{g'(x)}$
+    \end{center}
+
+    Dann ist $g(x) \neq 0$ für alle $x > a$, und es gilt
+
+    \begin{center}
+        \eqboxf{$\lim\limits_{x \to a^+} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to a^+} \dfrac{f'(x)}{g'(x)}$}
+    \end{center}
+
+    Existiert $\lim\limits_{x \to a^+} \frac{f'(x)}{g'(x)}$ nicht, so muss man nochmals l'Hôpital anwenden.
+
+
+    \subsection{Der Umkehrsatz}
+
+    Sei $f:]a,b[ \to \R$ differenzierbar mit $f'(x) > 0$ auf $]a,b[$. Seien
+
+    \begin{center}
+        $-\infty \leq c = \inf\limits_{a < x < b} f(x) < \sup\limits_{a < x < b} f(x) = d \leq \infty$
+    \end{center}
+
+    Dann ist $f:]a,b[\, \to \,]c,d[$ bijektiv und die Umkehrfunktion $f^{-1}:\,]c,d[\, \to \,]a,b[$ ist differenzierbar für $y \in ]c,d[$ und das Differential ist
+
+    \begin{center}
+        \eqboxf{$(f^{-1})' (y) = \dfrac{1}{f'(f^{-1}(y))}$}
+    \end{center}
+
+
+    \subsection{Funktionen der Klasse $C^1$}
+
+    Sei $\Omega$ ein offenes Intervall und $f: \Omega \to \R$ differenzierbar. $f$ heisst von der Klasse $C^1(\Omega, \R)$, falls die Ableitung $f'(x)$ stetig ist. Es gilt also
+
+    \begin{center}
+        \eqbox{$C^1(\Omega, \R) = \{ f: \Omega \to \R; f, f' \text{ stetig} \}$}
+    \end{center}
+
+    \subsubsection{Satz}
+
+    Sei $(f_k)_{k \in \mathbb{N}}$ eine Folge von Funktionen in $C^1(\Omega, \R)$ mit
+
+    \begin{center}
+        $f_k \xrightarrow[]{glm} f, f'_k \xrightarrow[]{glm} g \quad (k \to \infty)$
+    \end{center}
+
+    wobei $f, g: \Omega \to \R$. Dann gilt $f \in C^1(\Omega, \R)$ und $f' = g$.
+
+
+    \subsection{Höhere Ableitungen}
+
+    Sei $m \in \N$. $f$ heisst auf $\Omega$ m-Mal differenzierbar, falls die $m$-te Ableitung von $f$ existiert und wird folgendermassen notiert:
+
+    \begin{center}
+        $f^{(m)} = \dfrac{d^m f}{d x^m}$
+    \end{center}
+
+    \subsubsection{Funktionen der Klasse $C^m$}
+
+    Sei $\Omega$ ein offenes Intervall und $f: \Omega \to \R^n$ $m$-Mal differenzierbar. $f$ heisst von der Klasse $C^m(\Omega, \R)$, falls $f^m(x)$ stetig ist. Es gilt also
+
+    \begin{center}
+        \eqbox{$C^m(\Omega, \R) = \{ f: \Omega \to \R; f \text{ ist \emph{m}-mal diffbar, } f, \dots, f^{(m)} \text{ stetig} \}$}
+    \end{center}
+
+    Bemerkung: Falls $f \in C^m(\Omega, \R)$ für alle $m \in \mathbb{N}$, dann schreibt man $f \in C^\infty(\Omega, \R)$. Solche Funktionen nennt man ''Glatte Funktionen''.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Taylor Entwicklung}
+    %Mehr Infos: https://brilliant.org/wiki/taylor-series-error-bounds/
+
+    %Gutes Bild: Vorlesung 27.04.23 @11min
+
+    Sei $f \in C^{(m+1)}([a,b], \R)$ auf $]a,b[$ $m$-Mal differenzierbar. Die Taylorentwicklung $m$-ter Ordnung von $f$ am Entwicklungspunkt $a$ ist
+
+    \begin{center}
+        \eqboxf{$\displaystyle T_m f(x; a) = \sum\limits_{k = 0}^m f^{(k)} (a) \dfrac{(x - a)^k}{k!} + \underbrace{f^{(m+1)}(\xi) \dfrac{(x - a)^{m+1}}{(m+1)!}}_{\text{Restterm}}$}
+    \end{center}
+
+    Wobei $\xi \in \,]a,b[$ ist, d.h. eine beliebige Zahl im Definitionsbereich.
+
+    \subsubsection{Beste Approximation}
+
+    Je näher $x$ bei $a$ liegt, desto besser approximiert das Taylorpolynom $T_m f(x; a)$ an der Stelle $x$ die Funktion $f$. Für $a < x < b$ gilt
+
+    \begin{center}
+        \eqbox{$\lim\limits_{x \to a} \dfrac{f(x) - T_m f(x,a)}{(x - a)^m} = 0$}
+    \end{center}
+
+    \subsubsection{Abschätzung vom Restterm}
+
+    Der Restterm $r_m f(x;a)$ besitzt folgende Abschätzung:
+
+    \begin{center}
+        \eqbox{$|r_mf(x,a)| \leq \sup\limits_{a < \xi < x} |f^{(m+1)}(\xi)| \dfrac{(x - a)^{m + 1}}{(m + 1)!}$}
+    \end{center}
+
+
+    \subsection{Lokale Extrema}
+
+    Sei $\Omega \subset \R$ offen und $f: \Omega \to \R$. Ein $x_0 \in \Omega$ heisst (\emph{strikte}) lokale Minimalstelle von $f$, falls in einer Umgebung $U$ von $x_0$ gilt
+
+    \begin{center}
+        $f(x) \geq f(x_0), \, \forall x \in U$ \,\, (bzw. $f(x) > f(x_0), \, \forall x \in U\setminus \{x_0\}$)
+    \end{center}
+
+    Analoges gilt für lokale Maximastellen.
+
+    \subsubsection{Satz}
+
+    Sei $\Omega \subset \R$ offen, $f \in C^2(\Omega, \R)$ und $x_0 \in \Omega$. Es gilt folgendes:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \begin{tabular}{r l} \toprule
+            i)   & Wenn $f'(x_0) = 0 \, \Rightarrow x_0$ ist ein lokales Extrema.                  \\
+            ii)  & Sind $f'(x_0) = 0$ und $f''(x_0) > 0 \, \Rightarrow$ $x_0$ ein lokales Minimum. \\
+            iii) & Sind $f'(x_0) = 0$ und $f''(x_0) < 0 \, \Rightarrow$ $x_0$ ein lokales Maximum. \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \textbf{Achtung}: Randpunkte vom Definitionsbereich nicht vergessen!
+
+    \subsubsection{Allgemeinerer Satz}
+
+    Sei $f \in \C^n(\Omega, \R)$ mit $f'(x_0) = \dots = f^{(n-1)}(x_0) = 0$ und $f^{(n)}(x_0) \neq 0$.
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \begin{tabular}{r l} \toprule
+            i)   & Ist $n$ gerade sowie $f^{(n)}(x_0) > 0$ $\,\Rightarrow$ $x_0$ ein lokales Minimum. \\
+            ii)  & Ist $n$ gerade sowie $f^{(n)}(x_0) < 0$ $\,\Rightarrow$ $x_0$ ein lokales Maximum. \\
+            iii) & Ist $n$ ungerade, so hat $f$ bei $x_0$ einen Wendepunkt.                           \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+
+    \subsection{Konvexe Funktionen}
+
+    Sei $f \in C^2(]a,b[, \R)$ mit $f'' \geq 0$. Ein Funktion $f$ heisst \emph{konvex}, falls für alle $x_0, x_1 \in \, ]a,b[$ und für alle $t \in [0,1]$ gilt:
+
+    \begin{center}
+        \eqbox{$f(t \cdot x_1 + (1 - t)\cdot x_0) \leq t \cdot f(x_1) + (1 - t) \cdot f(x_0)$}
+    \end{center}
+
+    Die Funktionen für welche dies gilt heissen Konvex. \medskip
+
+    Bemerkung: Wenn $f''(x) > 0$ $\Rightarrow$ $f(x)$ ist konvex.
+    \vfill\null
+    \columnbreak
+
+
+    \pagebreak
+
+
+    \section{Integralrechnung auf $\R$}
+
+
+    \subsection{Stammfunktionen (SF)}
+
+    $F \in C^1(]a,b[)$ heisst Stammfunktion zu $f$, falls für alle $x \in ]a,b[$ gilt:
+
+    \begin{center}
+        \eqbox{$\displaystyle \int f(x) dx = F(x)$ \qquad $F'(x) = f(x)$}
+    \end{center}
+
+
+    \subsubsection{Satz: Integrationskonstante}
+
+    Zwei Stammfunktionen $F_1, F_2 \in C^1(]a,b[)$ einer Funktion unterscheiden sich nur durch eine Integrationskonstante voneinander: $F_1 - F_2 \equiv c \in \R$
+
+
+    \subsubsection{Das Integral}
+
+    Sei $F \in C^1(]a,b[,\R)$ eine SF von $f$. Das Integral von $f$ über $[a, b]$ ist:
+
+    \begin{center}
+        \eqbox{$\displaystyle \int\limits_{a}^b f(t) dt := F(b)-F(a)$}
+    \end{center}
+
+
+    \subsection{Eigenschaften vom Integral (und R-Integral)}
+
+    \subsubsection{Linearität}
+
+    Seien $f,g \in C^0(]a,b[)$ mit SFs $F,G \in C^1(]a,b[)$, und $\alpha, \beta \in \R$. Dann gilt
+
+    \begin{center}
+        \eqbox{$\displaystyle \int [\alpha \cdot f(x) + \beta \cdot g(x)] dx = \alpha \cdot \int f(x) dx + \beta \cdot \int  g(x) dx$}
+    \end{center}
+
+
+    \subsubsection{Monotonie}
+
+    Seien $f,g \in C^0(]a,b[)$ mit SFs $F,G \in C^1(]a,b[)$, und $f \leq g$. Dann gilt
+
+    \begin{center}
+        \eqbox{$\displaystyle \int\limits_{x_0}^{x_1} f(t) dt \leq \int\limits_{x_0}^{x_1} g(t) dt$}
+    \end{center}
+
+
+    \subsubsection{Gebietsadditivität}
+
+    Sei $f \in C^0(]a,b[)$ mit SF $F \in C^1(]a,b[)$. Für $a < x_0 \leq x_1 \leq x_2 < b$ gilt:
+
+    \begin{center}
+        \eqbox{$\displaystyle \int\limits_{x_0}^{x_1} f(x) dx + \int\limits_{x_1}^{x_2} f(x) dx = \int\limits_{x_0}^{x_2} f(x) dx$}
+    \end{center}
+
+    \subsubsection{Standardabschätzung}
+
+    Sei $f \in C^0([a,b])$. Dann gilt folgende Abschätzung
+
+    \begin{center}
+        \eqbox{$\displaystyle \left| \int\limits_a^b f(x) dx \right| \leq \int\limits_a^b |f(x)| dx \leq ||f(x)||_{C^0} (b - a)$}
+    \end{center}
+
+    \subsubsection{Korollar bezüglich glm. Konveregenz}
+
+    Seien $f, f_k \in C^0([a,b])$ mit $f_k \xrightarrow[]{glm} f \, (k \to \infty)$. Dann gilt
+
+    \begin{center}
+        \eqbox{\hspace*{-5pt}$\displaystyle \left| \int\limits_a^b f_k dx - \int\limits_a^b f dx \right| \leq \int\limits_a^b |f_k - f| dx \leq (b - a) \, || f_k - f||_{C^0} \to 0 (k \to \infty)$\hspace*{-5pt}}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Treppenfunktionen}
+
+    $f: [a,b] \to \R$ heisst Treppenfunktion, falls für eine Zerlegung $I = [a,b]$ in disjunkte (abgeschlossene, offene, halboffene) Teilintervalle $I_1, \dots, I_K$ mit dazugehörigen \emph{Konstanten} $c_k \in \R$ gilt:
+
+
+    \begin{minipage}{0.74\linewidth}
+        \begin{center}
+            \eqbox{$\displaystyle f(x) = \sum\limits_{k = 1}^K c_k \cdot \chi_{I_k}$ mit $\chi_{I_k}(x) = \begin{cases}
+                        1 & , x \in I_k \\ 0 & , x \not\in I_k \\
+                    \end{cases}$}
+        \end{center}
+    \end{minipage}
+    \begin{minipage}{0.25\linewidth}
+        \begin{center}
+            \includegraphics[width=1\linewidth]{Bilder/RiemannSumme.png}
+        \end{center}
+    \end{minipage}
+
+
+
+    Sei $|I_k|$ die Länge von $I_k$. Das Integral einer Treppenfunktion $f(x)$ ist
+
+    \begin{center}
+        \eqbox{$\displaystyle \int\limits_{a}^{b} \left(\sum\limits_{k = 1}^K c_k \cdot \chi_{I_k}\right) dx = \sum\limits_{k = 1}^K c_k \cdot |I_k|$}
+    \end{center}
+
+    \subsubsection{Lemma}
+
+    Sind $e,g:[a,b] \to \R$ Treppenfunktionen mit $e \leq g$, dann gilt
+
+    \begin{center}
+        $\int\limits_a^b e(x) dx \leq \int\limits_a^b g(x) dx$
+    \end{center}
+
+
+    \subsection{Die Riemannsche Summe}
+
+    Sei $f\in C^0([a,b])$. Dann gilt für eine beliebige Folge von disjunkte Zerlegung von $I$ in Teilintervalle $I_k^n, 1 \leq k \leq K_n$ mit \emph{Feinheit}
+
+    \begin{center}
+        $\delta_n = \sup\limits_{1 \leq k \leq K_n} |I_k^n| \to 0 \quad (n \to \infty)$
+    \end{center}
+
+    und eine beliebigen Auswahl an Punkten $x_k^n \in I^n_k, \, 1 \leq k \leq K_n$, stets
+
+    \begin{center}
+        \eqbox{$\displaystyle\int\limits_a^b \left( \sum\limits_{k = 1}^{K_n} f(x_k^n) \chi_{I_k^n}\right) dx = \sum\limits_{k = 1}^{K_n} f(x_k^n) | I_k^n | \to \int\limits_a^b f(x) dx \, (n \to \infty)$}
+    \end{center}
+
+
+    \subsection{Das Riemannsche Integral (R-Integral)}
+
+    Sei $f:[a,b] \to \R$ beschränkt und seien $e(x),g(x)$ Treppenfunktionen.
+
+    \begin{center}
+        Untere Riemann-Integral von $f$: \, \eqbox{$\displaystyle \underline{\int\limits_{a}^b} f(x) dx = \sup\limits_{e(x) \leq f(x)} \int\limits_a^b e(x) dx$} \medskip
+
+        Obere Riemann-Integral von $f$: \eqbox{$\displaystyle \overline{\int\limits_a^b} f(x) dx = \inf\limits_{g(x) \geq f(x)} \int\limits_a^b g(x) dx$}
+    \end{center}
+
+    Ein solches $f(x)$ heisst über $[a,b]$ Riemann-integrabel, falls
+
+    \begin{minipage}{0.3\linewidth}
+        \begin{center}
+            \includegraphics[width=0.8\linewidth]{Bilder/R-Integral.png}
+        \end{center}
+    \end{minipage}
+    \begin{minipage}{0.69\linewidth}
+        \begin{center}
+            \eqbox{$\displaystyle \underline{\int\limits_{a}^b} f(x) dx = \overline{\int\limits_a^b} f(x) dx := \int\limits_a^b f(x) dx$}
+        \end{center}
+    \end{minipage}
+
+    \subsubsection{Sätze}
+
+    i) $f:[a,b] \to \R$ monoton, beschränkt $\Rightarrow f$ ist über $[a,b]$ R-integrabel. \medskip
+
+    ii) $f:[a,b] \to \R$ stetig ($f \in C^0([a,b], \R)$) $\Rightarrow f$ ist über $[a,b]$ R-integrabel.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Substitutionsregel}
+
+    Seien $f,g \in C^1(]a,b[)$. Dann gilt für $a < x_0 < x_1 < b$:
+
+    \begin{center}
+        \eqboxf{$\displaystyle \int\limits_{x_0}^{x_1} f'(\underbrace{g(x)}_{= u}) \cdot \underbrace{g'(x) dx}_{= d u} = \int\limits_{g(x_0)}^{g(x_1)} f'(u) du$}
+    \end{center}
+
+
+    \subsection{Partielle Integration}
+
+    Seien $u,v \in C^1(]a,b[)$, so dass $u(x) \cdot v'(x)$ eine SF besitzt. Dann besitzt $u'(x) \cdot v(x)$ auch eine SF und es gilt
+
+    \begin{center}
+        \eqboxf{$\displaystyle \int\limits_a^b u'(x) v(x) dx = [u(x)v(x)]_a^b - \int\limits_a^b u(x) v'(x) dx $}
+    \end{center}
+
+    \subsubsection{Bei periodischen Funktion}
+
+    Bei Partieller Integration von zwei periodischen Funktionen geht man eine Periode durch und sotiert dann das ''ursprüngliche Integral'' auf die Linke Seite und kann so, dass Integral berechnen.
+
+
+    \subsection{Hauptsatz der Differential- und Integralrechnung}
+
+    Sei $f \in C^0([a,b], \R)$. So ist für jedes $c \in [a,b]$ die Integralfunktion
+
+    \begin{center}
+        \eqbox{$\displaystyle F: [a,b] \to \R$ mit $\displaystyle F(x) = \int\limits_c^x f(t) dt$}
+    \end{center}
+
+    differenzierbar und es gilt für alle $x \in [a,b]$: $F'(x) = f(x)$.
+
+    \subsubsection{Anwendung: Parameterintegral}
+
+    \begin{center}
+        $\displaystyle \frac{d}{dt} \int\limits_a^{h(t)} g(x) dx = \frac{d}{dt} [G(x)]^{h(t)}_{a} = g(h(t)) \cdot h'(t)$
+    \end{center}
+
+
+    \subsection{Uneigentliches Riemann-Integral}
+
+    Sei $f: ]a,b[ \to \R$ mit $-\infty \leq a < b \leq \infty$ über jedes kompakte Intervall $[c,d] \subset ]a,b[$ R-integrabel. $f$ ist über $]a,b[$ \emph{uneigentlich R-integrabel}, falls folgender Grenzwert existiert:
+
+    \begin{center}
+        \eqbox{$\displaystyle \int\limits_{a}^b f(x) dx = \lim\limits_{c \to a^+} \int\limits_{c}^0 f(x) dx + \lim\limits_{d \to b^-} \int\limits_{0}^d f(x) dx$}
+    \end{center}
+
+    \subsubsection{Satz: Reihenkonvergenz}
+
+    Sei $f:[1, \infty) \to \R_+$ \emph{monoton fallend}. Dann gilt
+
+    \begin{center}
+        \eqbox{$\displaystyle \sum\limits_{k = 1}^\infty f(x)$ konvergiert $\Leftrightarrow$ $\displaystyle \int\limits_1^\infty f(x) dx$ konvergiert}
+    \end{center}
+    \vfill\null
+\end{multicols*}
+
+\begin{multicols*}{3}
+    \section{Gewöhnliche lineare Differentialgleichungen (GDG)}
+
+
+    \subsection{Differentialgleichungen 1ter Ordnung}
+
+    \subsubsection{Homogene Lösung}
+
+    Die Homogene Differentialgleichung 1ter Ordnung hat die Form:
+
+    \begin{center}
+        \eqbox{$y'(x) - a(x) y(x) = 0$}
+    \end{center}
+
+    Den Lösungsansatz nennt man \emph{Seperation der Variablen}:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{r p{0.8\linewidth}} \toprule
+            i)   & $y'(x) - a(x) y(x) = 0 \Leftrightarrow \dfrac{d y}{d x} = a(x) y(x)$                     \\
+            ii)  & Umformen auf folgende Form:
+
+            \begin{center}
+                $\frac{1}{y(x)} dy = a(x) dx$
+            \end{center}                                                           \\
+            iii) & Auf beiden Seiten integrieren:
+
+            \begin{center}
+                $\displaystyle \int \frac{1}{y(x)} dy = \int a(x) dx \Rightarrow \log(y(x)) = A(x) + C$
+            \end{center} \\
+            iv)  & Auf beide Seiten $e^x$ anwenden:
+
+            \begin{center}
+                \eqboxf{$y_h(x) = e^{A(x)} \cdot C$}
+            \end{center}                                       \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    Bemerkung: $A(x)$ ist die SF von $a(x)$ und $C$ die Integrationskonstante.
+
+
+    \subsubsection{Inhomogene Differentialgleichungen 1ter Ordnung}
+
+    Eine Inhomogene Differentialgleichung 1ter Ordnung hat die Form:
+
+    \begin{center}
+        \eqbox{$y'(x) - a(x) y(x) = b(x)$}
+    \end{center}
+
+    Den Lösungsansatz nennt man \emph{Variation der Konstanten}:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{r p{0.8\linewidth}} \toprule
+            i)   & Die homogene Lösung $y'(x) - a(x) y(x) = 0$ berechnen.                                                                  \\
+            ii)  & Man macht den Ansatz, dass $C$ von $x$ abhängt:
+
+            \begin{center}
+                $y_h(x) = C(x) \cdot e^{A(x)}$
+            \end{center}                                                                            \\
+            iii) & Man berechnet $y_h'(x)$:
+
+            \begin{center}
+                $y_h'(x) = C'(x)e^{A(x)} + C(x) a(x) e^{A(x)}$
+            \end{center}                                      \\
+            iv)  & $y_h'(x)$ und $y_h(x)$ in die ursprüngliche DGL einsetzen:
+
+            \begin{center}
+                $C'(x)e^{A(x)} \underbrace{+ C(x) a(x) e^{A(x)} - C(x) a(x) e^{A(x)}}_{= 0} = b(x)$
+            \end{center}                                                            \\
+            v)   & Gleichung umstellen und dann Integrieren:
+
+            \begin{center}
+                $\displaystyle C'(x) = \dfrac{b(x)}{e^{A(x)}} \Rightarrow C(x) = \int \dfrac{b(x)}{e^{A(x)}} dx + K$
+            \end{center}                          \\
+            vi)  & Gefundes $C(x)$ in $y_h(x) = C(x) \cdot e^{A(x)}$ einsetzen, dies ist die Lösung der inhomogenen Differentialgleichung.
+            \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Homogene Systeme linearer Differentialgleichungen}
+
+    Sei $F(t) \in \R^n$ und $A \in M_{n \times n}(\R)$. Folgende Gleichung
+
+    \begin{center}
+        \eqbox{$\dfrac{d F(t)}{d t} = A \cdot F(t)$}
+    \end{center}
+
+    ist die Standardform eines homogenen Systems linearer Differentialgleichungen mit konstanten Koeffizienten.
+
+    \subsubsection{Existenz- und Eindeutigkeitssatz}
+
+    Sei $F(t), F_0 \in \R^n$ und $A \in M_{n \times n}(\R)$. Das folgende \emph{Anfangswertproblem}
+
+    \begin{center}
+        \eqbox{$\dfrac{d F}{d t} = A \cdot F(t), \quad F(0) = F_0$}
+    \end{center}
+
+    besitzt \emph{genau} eine Lösung $F \in C^1(\R; \R^n)$.
+
+
+    \subsubsection{Die Fundamentallösung}
+
+    Folgende Matrix-wertige Funktion $\Phi(t) \in C^1(\R, M_{n \times n} (\R))$
+
+    \begin{center}
+        \eqbox{$t \mapsto \Phi(t) := Exp(A t) = \sum\limits_{k = 0}^\infty \dfrac{A^k t^k}{k!}$}
+    \end{center}
+
+    besitzt die erwünschten Eigenschaften
+
+    \begin{center}
+        $\dfrac{d\Phi}{d} = A \cdot \Phi(t), \, \Phi(0) = id$
+    \end{center}
+
+    Sie heisst Fundamentallösung von dem System $\dot F = A F(t)$.
+
+
+    \subsubsection{Die Fundamentallösung einer diagonalisierbaren Matrix}
+
+    Sei $A$ diagonalisierbar ($A$ ist ähnlich zu einer Diagonalmatrix), d.h.:
+
+    \begin{center}
+        $A = T \begin{pmatrix}
+                \lambda_1 &        & 0         \\
+                          & \ddots &           \\
+                0         &        & \lambda_n \\
+            \end{pmatrix} T^{-1}$ \qquad \eqboxf{$\det(A - \lambda I) = 0$}
+    \end{center}
+
+    Wobei $T$ die $n$ Eigenvektoren von $A$ als Spaltenvektoren enthält. \medskip
+
+    In einem solchen Fall nimmt die Fundamentallösung folgende Form an:
+
+    \begin{center}
+        \eqbox{$\Phi(t) = Exp(t A) = T \begin{pmatrix}
+                    e^{t \lambda_1} &        & 0               \\
+                                    & \ddots &                 \\
+                    0               &        & e^{t \lambda_n} \\
+                \end{pmatrix} T^{-1}$}
+    \end{center}
+
+    Wir wissen ausserdem, dass alle Lösungen von folgender Form sind:
+
+    \begin{center}
+        $\displaystyle F(t) = \sum\limits_{i = 1}^n B_i \cdot e^{\lambda_i t}$
+    \end{center}
+
+    wobei $B_i \in \R^n$.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Reduktion der Ordnung}
+
+    Die Allgemeine Form einer DGL n-ter Ordnung ist:
+
+    \begin{center}
+        \eqbox{$f^{(n)} + a_{n-1} f^{(n-1)} + \dots +a_1 \dot f + a_0 f = 0$}
+    \end{center}
+
+    Die Reduktion zu einem System 1.Ordnung ist:
+
+    \begin{center}
+        $\underbrace{\begin{pmatrix}
+                    \dot f \\ \ddot f \\ \vdots \\ f^{(n)} \\
+                \end{pmatrix}}_{\dot F} = \underbrace{\begin{bmatrix}
+                    0      & 1    &        & 0        \\
+                    \vdots &      & \ddots &          \\
+                    0      & 0    &        & 1 \quad  \\
+                    -a_0   & -a_1 & \dots  & -a_{n-1} \\
+                \end{bmatrix}}_{A_{n \times n}} \underbrace{\begin{pmatrix}
+                    f \\ \dot f \\ \vdots \\ f^{(n-1)} \\
+                \end{pmatrix}}_{F(t)}$
+    \end{center}
+
+
+    \subsection{Der Exponentialansatz}
+
+    Wir wissen dank der Fundamentallösung, dass man Lösungen der Form
+
+    \begin{center}
+        \eqbox{$f(t) = e^{\lambda t}$} \quad (wobei $\dot f = \lambda e^{\lambda t}, \dots, \, f^{(n)}(t) = \lambda^n e^{\lambda t}$)
+    \end{center}
+
+    sucht. Setzt man diesen Lösungsansatz in die allgemeine Form ein
+
+    \begin{center}
+        \eqboxf{$\underbrace{\lambda^n + a_{n-1} \lambda^{n-1} + \dots + a_1 \lambda + a_0}_{p(\lambda)} = 0$}
+    \end{center}
+
+    dann kriegt man das \emph{charakteristische Polynom} $p(\lambda)$ der linearen GDG.
+
+
+    \subsubsection{Beziehung zum charakteristischen Polynomen der Matrix $A$}
+
+    Die Beziehung zum charakteristischen Polynom der Matrix $A_{n \times n}$ ist:
+
+    \begin{center}
+        \eqbox{$\chi_A ( \lambda):= \det(A - \lambda I_n) = (-1)^n p(\lambda)$}
+    \end{center}
+
+
+    \subsection{Der Lösungsraum}
+
+    Sei $A \in M_{n \times n}(\R)$. Der Lösungsraum
+
+    \begin{center}
+        \eqbox{$X_A = \{F \in C^1(\R, \R^n) ; \, \dot F = A F\}$}
+    \end{center}
+
+    bildet ein $n$-dimensionaler $\R$-Vektorraum (Unterraum von $C^1(\R, \R^n)$). \medskip
+
+    Analoges gilt für $A \in M_{n \times n}(\C)$:
+
+    \begin{center}
+        $\tilde{X}_A = \{F \in C^1(\R, \C^n) ; \, \dot F = A F\}$
+    \end{center}
+
+    bildet ein $n$-dimensionaler $\C$-Vektorraum.
+
+    \subsubsection{Korollar}
+
+    Sei $a = \{a_0, a_1, \dots, a_{n-1}\} \in \R$ (oder $\C$). Der Lösungsraum
+
+    \begin{center}
+        $Z_a = \{f \in C^n (\R, \R^n) ; \, f^{(n)} + a_{n-1} f^{(n-1)} + \dots + a_0 f = 0\}$
+    \end{center}
+
+    ist ein $n$-dimensionaler Unterraum von $C^n(\R, \R^n)$.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Einschub: Der Fundamentalsatz der Algebra}
+
+    Sei $\lambda_i \in \mathbb{C}$ und $m_i \in \mathbb{N}$. Ein Polynom besitzt $n$ Nullstellen in $\C$, d.h.:
+
+    \begin{center}
+        $\displaystyle p(\lambda) = \prod\limits_{i = 1}^{l} (\lambda - \lambda_i)^{m_i}$ \qquad wobei \,\, $\displaystyle \sum\limits_{i = 1}^l m_i = n$
+    \end{center}
+
+    wobei $\lambda_i$ die Nullstellen mit jeweiliger Vielfachheit $m_i$ sind.
+
+
+    \subsection{Ableitungsoperator und Identitätsoperator}
+
+    Die Ableitungsoperator ist eine lineare Abbildung:
+
+    \begin{center}
+        \eqbox{$D: C^{\infty}(\R, \R) \to C^{\infty}(\R, \R) \qquad f \mapsto \dot f = D f$}
+    \end{center}
+
+    Bem: Der Identitätsoperator ($Id(f) = f$) ist offensichtlich auch linear. \medskip
+
+    Man kann das char. Polynoms $p(\lambda)$ auch, wie folgt, beschreiben:
+
+    \begin{center}
+        \eqbox{$\displaystyle p(D) = \prod\limits_{i = 1}^{l} (D - \lambda_i \cdot Id)^{m_i}$}
+    \end{center}
+
+    \subsubsection{Beispiel}
+
+    \begin{center}
+        $p(\lambda) = (\lambda - 1)^2 \,\, \Rightarrow p(D)f = (D-Id)^2f = (D - Id)(\dot f - f) = \ddot f - 2 \dot f + f$
+    \end{center}
+
+
+    \subsection{Der Hauptsatz vom Kapitel}
+
+    Es sei eine DGL n-ter Ordnung in der allgemeinen Form:
+
+    \begin{center}
+        $f^{(n)} + a_{n-1} f^{(n-1)} + \dots +a_1 \dot f + a_0 f = 0$
+    \end{center}
+
+    Dann ist, gemäss Exponentialansatz, ihr charakteristisches Polynom:
+
+    \begin{center}
+        \eqbox{$\displaystyle p(\lambda) = \lambda^n + a_{n-1} \lambda^{n-1} + \dots + a_1 \lambda + a_0 = \prod\limits_{i = 1}^{l} (\lambda - \lambda_i)^{m_i}= 0$}
+    \end{center}
+
+    Jede Lösung der Differentialgleichung ist darstellbar als Linearkombination folgender $n$ linear unabhängigen Funktionen:
+
+    \begin{center}
+        \eqboxf{$\displaystyle f_{ik} (t) = t^k e^{\lambda_i t}$ wobei $0 \leq k < m_i$} \quad $\displaystyle \Rightarrow f(t) = \sum\limits_{i = 1}^l  C_i \cdot f_{ik}$
+    \end{center}
+
+    wobei die Koeffizienten $C_i \in \R$ durch die zur GDG dazugehörigen Anfangswerte $f(0), \dot f(0), \dots, f^{(n-1)}(0)$ bestimmt werden. \medskip
+
+    Bem: Diese Linearkombination bildet eine Basis vom Lösungsraum $Z_a$.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Inhomogene Differentialgleichungen höherer Ordnung}
+
+    Seien $A \in M_{n \times n}(\R)$ und $B \in C^{0}(\R, \R^n)$. Sei $F_{part} \in C^1(\R,\R^n)$ eine beliebige ''\emph{partikuläre}'' Lösung von
+
+    \begin{center}
+        \eqbox{$\dot F = A F(t) + B(t)$}
+    \end{center}
+
+    Dann ist jede dazugehörige Lösung $F$ von der Form
+
+    \begin{center}
+        \eqboxf{$F(t) = F_{part}(t) + F_{hom} (t)$}
+    \end{center}
+
+    wobei $F_{hom}$ eine Lösung der homogenen Gleichung $\dot F = A F(t)$ ist. \medskip
+
+    Eindeutigkeitssatz: Insbesondere gibt es zu jedem $F_0 \in \R^n$ stets genau eine Lösung $F(t)$ mit $F(0) = F_0$.
+
+    \subsubsection{Allgemeines Vorgehen zur Berechnung der partikulären Lösung}
+
+    Die partikuläre Lösung $F_{part}(t) \in C^1(\R,\R^n)$ löst folgende Gleichung:
+
+    \begin{center}
+        \eqbox{$\dot F_{part} = A F_{part} (t) + B(t)$ \qquad bzw. $f^{(n)} + \dots + a_0 f = b(t)$}
+    \end{center}
+
+    Das Vorgehen ist, wie folgt:
+
+    \begin{center}
+        \begin{tabular}{r p{0.8\linewidth}} \toprule
+            i)   & Den der Störfunktion $b(t)$ entsprechenden Ansatz suchen. \\
+            ii)  & Ableitungen des Ansatzes $f_{part}$ berechnen.            \\
+            iii) & $f_{part}$ mit ihren Ableitungen in die GDG einsetzen.    \\
+            iv)  & Die Koeffizienten vom Lösungsansatz durch einen Koeffizientenvergleich bestimmen. \\ 
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \begin{center}
+        \includegraphics[width = 1\linewidth]{Bilder/DGL_Partikulare.jpeg}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Kochrezept: Vorgehen bei DGLs}
+
+    \begin{enumerate}
+        \item Homogene Lösung bestimmen.
+        \begin{itemize}
+            \item Charakteristisches Polynom bestimmen (Exponentialansatz)
+            \item Nullstellen bestimmen und im Hauptsatz einsetzen. 
+            \item Komplex konjugierte imaginäre Nullstellen ersetzen durch das entsprechende cos/sin Paar.
+        \end{itemize} 
+        \item Partikuläre Lösung bestimmen, falls eine Störfunktion vorhanden ist.
+        \item Anfangswertproblem auflösen.
+    \end{enumerate}
+
+    \subsection{Sonstiges}
+
+    \subsubsection{Harmonische Oszillatoren}
+
+    Harmonische Oszillatoren besitzen folgende DGL:
+
+    \begin{center}
+        $\ddot f + \omega_0^2 f = 0, \, \omega^2_0 > 0 \quad \Rightarrow p(\lambda) = \lambda^2 + \omega_0 = 0 \quad \Rightarrow \lambda_1,2 = \pm i \omega_0$
+    \end{center}
+
+    Die Allgemeine Lösung hat, unter anderem, folgende Formen:
+
+    \begin{center}
+        $f(t) = A e^{i \omega_0 t} + B e^{-i \omega_0 t}$ \qquad $f(t) = A \cos(\omega_0 t) + B \sin(\omega_0 t)$
+    \end{center}
+
+    Die Reelle Lösung (Physikalische) bildet man durch die Summe und Differenz der ersten Lösung unter Anwendung der Eulerschen Formel (Koeffizienten A,B erst am Schluss anfügen).
+
+
+    \subsubsection{Erzwungene Schwingungen}
+
+    Die folgende DGL einer erzwungene Schwingung wandelt man, wie folgt, zur Berechnung der partikulären:
+
+    \begin{center}
+        $\ddot f + 2\delta \dot f + \omega_0^2 f = \beta_0 \cos(\omega t) \quad \Rightarrow \ddot f + 2\delta \dot f + \omega_0^2 f = \beta_0 e^{i\omega t}$
+    \end{center}
+
+    Benutzt man den Ansatz $f_{part} = c \cdot e^{i \omega t}$. So bekommt man:
+
+    \begin{center}
+        $c = \dfrac{\beta_0}{(\omega_0^2 - \omega^2) + 2 i \delta \omega} = \beta_0 \dfrac{\omega_0^2 - \omega^2 - 2i \delta \omega}{(\omega_0^2 - \omega^2)^2 + 4 \delta^2 \omega^2}$
+    \end{center}
+
+    Drückt man dies in der Polarform $R \cdot e^{i \varphi}$ aus:
+
+    \begin{center}
+        $R = \dfrac{\beta_0}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4 \delta^2 \omega^2}}$ \qquad $\varphi = \arctan\left( \dfrac{-2 \delta \omega}{\omega_0^2 - \omega^2} \right) \in \, ]-\pi,0[$
+    \end{center}
+
+    Und setzt $c$ wieder ein in $f_{part}$:
+
+    \begin{center}
+        $f_{part} = c \cdot e^{i \omega t} = R \cdot e^{i(\omega t + \varphi)}$
+    \end{center}
+
+    Nun kann man noch den Realteil nehmen und hat dann die gesuchte partikuläre Lösung:
+
+    \begin{center}
+        $\tilde{f}_{part}(t) = \Re{f_{part}(t)} = R \cdot \cos(\omega t + \varphi)$
+    \end{center}
+    \vfill\null
+    \pagebreak
+
+
+    \section{Differentialrechnung in $\R^n$}
+
+    \subsection{Partielle Ableitung}
+
+    Sei $\Omega \subset \R^n$ offen. $f: \Omega \to \R$ heisst in $x_0$ in Richtung $e_i = (0, \dots, 1, \dots 0)$ partiell differenzierbar, falls folgender Grenzwert exisitert:
+
+    \begin{center}
+        \eqboxf{$\dfrac{\partial f}{\partial x^i} (x_0) = \partial_{x^i} f (x_0) = \lim\limits_{h \to 0} \dfrac{f(x_0 + h \cdot e_i) - f(x_0)}{h}$}
+    \end{center}
+
+    \begin{center}
+        \includegraphics[width = 1\linewidth]{Bilder/PartiellAbleitung.png}
+    \end{center}
+
+
+    \subsubsection{Tangentialebene}
+    %https://de.wikipedia.org/wiki/Tangentialebene#Tangentialebene_an_den_Graphen_einer_Funktion
+
+    Die Tangentialebene ist die beste Approximation einer 2D-Funktion in der Nähe von $(x_0, y_0)$. Sie ist, wie folgt, definiert:
+
+    \begin{center}
+        \eqbox{$g(x,y) = f(x_0,y_0) + \dfrac{\partial f}{\partial x}(x_0,y_0) \cdot (x - x_0) + \dfrac{\partial f}{\partial y}(x_0,y_0) \cdot (y - y_0)$}
+    \end{center}
+
+
+    \subsection{Differential}
+
+    Sei $\Omega \subset \R^n$ offen. $f: \Omega \to \R$ heisst differenzierbar an der Stelle $x_0$, falls eine lineare Abbildung $A: \R^n \to \R$ existiert mit
+
+    \begin{center}
+        \eqbox{$\lim\limits_{x \to x_0} \dfrac{f(x) - f(x_0) - A \cdot (x - x_0)}{|| x - x_0||} = 0$}
+    \end{center}
+
+    Dann heisst $d f_{x_0} := A$ (ein sogenannter co-Vektor) das Differential von $f$ an der Stelle $x_0$ und desweiteren gilt
+
+    \begin{center}
+        $d f_{x_0} \cdot (x - x_0) = \begin{bmatrix}
+                \dfrac{\partial f}{\partial x^{0}} (x_0), \dots, \dfrac{\partial f}{\partial x^{n}} (x_0) \\
+            \end{bmatrix} \begin{pmatrix}
+                x^1 - x_0^1 \\ \vdots \\ x^n - x^n_0 \\
+            \end{pmatrix}$
+    \end{center}
+
+    Bem: $dx^{i} = (0, \dots, 0, 1, 0, \dots 0)$ sind die Basiselemente von $d f_{x_0}$.
+
+    \subsubsection{Satz: Kriterium für $C^1$}
+
+    Sei $\Omega \subset \R^n$ offen. $f: \Omega \to \R$ heisst von der Klasse $C^1$, $f \in C^1(\Omega)$ falls:
+
+    \begin{center}
+        \eqbox{\begin{tabular}{r p{0.8\linewidth}}
+                i)  & \hspace*{-10pt} $f$ ist an jeder Stelle $x_0 \in \Omega$ in jede Richtung $\vect{e}_i$ partiell differenzierbar.   \\
+                ii) & \hspace*{-10pt} Die Funktionen $x \mapsto \dfrac{\partial f}{\partial x^{i}} (x)$ sind auf $\Omega$ \emph{stetig}. \\
+            \end{tabular}}
+    \end{center}
+
+
+    \subsubsection{Satz}
+
+    $f \in C^1(\Omega) \Rightarrow$ $f$ ist an jeder Stelle $x_0$ \emph{differenzierbar und stetig} auf $\Omega$.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Differentiationsregeln}
+
+    Seien $f,g: \Omega \to \R$ an der Stelle $x_0 \in \Omega$ differenzierbar. Dann gilt
+
+    \begin{center}
+        \renewcommand{\arraystretch}{2}
+        \begin{tabular}{r l} \toprule
+            Summenregel:   & \hspace*{-10pt} $d(f + g)_{(x_0)} = df_{(x_0)} + dg_{(x_0)}$                                                           \\
+            Produktregel:  & \hspace*{-10pt} $d(f \cdot g)_{(x_0)} = df_{(x_0)} \cdot g(x_0) + f(x_0) \cdot dg_{(x_0)}$                             \\
+            Quotientregel: & \hspace*{-10pt} $d\left(\dfrac{f}{g}\right)_{(x_0)} = \dfrac{df_{(x_0)} \cdot g(x_0) - f(x_0) dg_{(x_0)}}{[g(x_0)]^2}$ \\ \bottomrule
+        \end{tabular}
+    \end{center}
+
+    Anwendung der Produktregel: $d f^n = n \cdot f^{n - 1} \cdot df$
+
+
+    \subsubsection{Satz: Kettenregel 1. Version}
+
+    Sei $g: \Omega \subset \R^n \to \R$ an der Stelle $x_0 \in \Omega$ differenzierbar und sei $f: \R \to \R$ an der Stelle $g(x_0)$ differenzierbar. Dann gilt
+
+    \begin{center}
+        \begin{minipage}{0.55\linewidth}
+            \begin{center}
+                \eqbox{$\underbrace{d(g \circ f)_{(x_0)}}_{\text{co-Vektor}} = f'(g(x_0)) \cdot \underbrace{d g_{(x_0)}}_{\text{co-Vektor}}$}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.44\linewidth}
+            \begin{center}
+                \includegraphics[width = 1\linewidth]{Bilder/Kettenregel_0.png}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+
+    \subsubsection{Satz: Kettenregel 2. Version}
+
+    Sei $g: ]a,b[ \to \Omega$ an der Stelle $t_0 \in ]a,b[$ differenzierbar und sei $f: \Omega \to \R$ $\Omega \subset \R^n$ an der Stelle $g(t_0)$ differenzierbar. Dann gilt
+
+    \begin{center}
+        \begin{minipage}{0.5\linewidth}
+            \begin{center}
+                \eqbox{$(f \circ g)' (t_0) = \underbrace{df_{(g(t_0))}}_{\text{co-Vektor}} \cdot \underbrace{g'(t_0)}_{\text{Vektor}}$}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.49\linewidth}
+            \begin{center}
+                \includegraphics[width = 1\linewidth]{Bilder/Kettenregel.png}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+
+    \subsection{Richtungsableitungen}
+
+    Sei $\Omega \subseteq \R^n$ offen und $f: \Omega \to \R$. Dann ist die Richtungsableitung eines Vektors $\vect{v} \in \R^n$ an einem Punkt $x_0$:
+
+    \begin{center}
+        \eqbox{$\partial_{\vect{v}} f(x_0) = \lim\limits_{s \to 0} \dfrac{f(x_0 + s \cdot \vect v) - f(x_0)}{s}$}
+    \end{center}
+
+    \subsubsection{Satz}
+
+    $f$ ist differenzierbar in $x_0 \in \Omega \, \Rightarrow \partial_v f(x_0) = df_{(x_0)} \cdot \vect{v}$
+
+
+    \subsection{Vektorfelder}
+
+    Funktionen der Form $v: \Omega \subset \R^n \to \R^n$ heissen Vektorfelder. Jeder Stelle im Definitionsbereich wird ein Vektor zugeordnet.
+
+
+    \subsubsection{Gradientenfeld}
+
+    Sei $f \in C^1(\R^n)$ mit der dazugehörigen 1-Form $\lambda = df$. Das zur 1-Form zugehörige Vektorfeld heisst Gradientenfeld und ist
+
+    \begin{center}
+        \eqbox{$\displaystyle \nabla f(\vect{x}) = \sum\limits_{i = 1}^n \dfrac{\partial f}{\partial x^i}(\vect{x}) \cdot \vect{e}^i = \begin{pmatrix}
+                    \frac{\partial f}{\partial x^1} \\ \vdots \\ \frac{\partial f}{\partial x^n}
+                \end{pmatrix}$}
+    \end{center}
+
+    \begin{center}
+        \begin{tabular}{r p{0.8\linewidth}} \toprule
+            i)  & \hspace*{-10pt} $\nabla f(\vect{x}_0)$ gibt die Richtung des ''steilsten Anstiegs'' an. \\
+            ii) & \hspace*{-10pt} $\nabla f(\vect{x}_0)$ ist orthogonal zu den ''Levelmengen''.           \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+
+    \subsection{Höhere Ableitungen}
+
+    Sei $\Omega \subset \R^n$ offen und $f \in C^1(\Omega)$. Dann ist $f \in C^2(\Omega)$, falls
+
+    \begin{center}
+        \eqbox{$\dfrac{\partial f}{\partial x^{i}} \in C^1(\Omega), \quad \forall 1 \leq i \leq n$}
+    \end{center}
+
+    D.h. falls \textbf{alle} zweiten part. Ableitungen existieren und stetig auf $\Omega$ sind.
+
+    \subsubsection{Satz von Hermann Schwarz}
+
+    Sei $\Omega \subset \R^n$ offen und $f \in C^2(\Omega)$. Dann gilt
+
+    \begin{center}
+        \eqbox{$\dfrac{\partial}{\partial x^{i}}\left(\dfrac{\partial f}{\partial x^{j}}\right) = \dfrac{\partial}{\partial x^{j}} \left(\dfrac{\partial f}{\partial x^{i}} \right), \quad \forall \, 1 \leq i, j \leq n$}
+    \end{center}
+
+
+    \subsection{Funktionen der Klasse $C^m$}
+
+    Sei $\Omega \subset \R^n$. $f \in C^1(\Omega)$ heisst von der Klasse $C^m$, $f \in C^m(\Omega)$, falls
+
+    \begin{center}
+        \eqbox{$\dfrac{\partial f}{\partial x^{i}} \in C^{m - 1}(\Omega), \quad \forall 1 \leq i \leq n$}
+    \end{center}
+
+    \subsubsection{Notation (Multi-Index Schreibweise)}
+
+    Sei $\alpha = (\alpha_1, \dots, \alpha_n) \in \N^n_0$ mit $|\alpha| = \alpha_1 + \dots + \alpha_n$ und $\alpha! = \alpha_1! \cdot \dots \cdot \alpha_n!$. Dann gelten für $x = (x_1, \dots, x_n)$ folgende Notationen:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{r l} \toprule
+            i)   & \hspace*{-10pt} $x^\alpha = x_1^{\alpha} \cdot \, \dots \, \cdot x_n^{\alpha_n}$                                                         \\
+            ii)  & \hspace*{-10pt} $\partial^{\alpha} f(x) = \dfrac{\partial^{|\alpha|} f(x)}{(\partial x_{1})^{\alpha_1} \dots (\partial x_n)^{\alpha_n}}$ \\
+            iii) & \hspace*{-10pt} $p(x) = \sum\limits_{|\alpha| \leq N} a_{\alpha} x^{\alpha}$                                                             \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+
+    \subsection{Der Satz von Taylor}
+
+    Sei $f \in C^m(\Omega \subset \R^n)$. Für jedes $x, y \in \R^n$ existiert ein $c \in [x, y]$, so dass
+
+    \begin{center}
+        \eqboxf{$f(y) = \underbrace{\sum\limits_{|\alpha| \leq (m - 1)} \dfrac{1}{\alpha!} \partial^{\alpha}f(x) (y - x)^{\alpha}}_{=: T_{m - 1}f(y, x)} + \underbrace{\sum\limits_{|\alpha| = m} \dfrac{1}{\alpha!} \partial^{\alpha}f(c) (y - x)^{\alpha}}_{\text{Restterm}}$}
+    \end{center}
+
+    $T_k f(y,x)$ heisst das Taylor-Polynom $k-ter$ Ordnung von $f(y)$ mit dem Entwicklungspunkt $x$.
+
+    \subsubsection{Taylorentwicklung für $n = 2$ und $m = 2$ }
+
+    \begin{center}
+        \renewcommand{\arraystretch}{2}
+        \begin{tabular}{r l}
+            \hspace*{-8pt} $f(y) = $ & \hspace*{-13pt} $f(x) + \frac{\partial f}{\partial x}(x) \cdot (y_1 - x_1) + \frac{\partial f}{\partial y}(x) \cdot (y_2 - x_2)$                  \\
+                                     & \hspace*{-13pt} $+ \frac{1}{2!} \, [ \frac{\partial^2 f}{\partial x_1^2}(c) (y_1 - x_1)^2 + \frac{\partial^2 f}{\partial x_2^2}(c) (y_2 - x_2)^2$ \\
+                                     & $ +2 \frac{\partial^2 f}{\partial x_2 \partial x_1}(c) (y_2 - x_2)(y_1 - x_1) ]$                                                                  \\
+        \end{tabular}
+    \end{center}
+
+    \subsubsection{Korollar}
+
+    Die Taylorentwicklung gibt die beste Approximation um $x$:
+
+    \begin{center}
+        \eqbox{$\lim\limits_{y \to x} \dfrac{f(y) - T_m f(y,x)}{|| y - x||^m} = 0$}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Hesse-Matrix}
+
+    Sei $\Omega \subset \R^n$ und $f \in C^2(\Omega)$. Die Hesse-Matrix von $f$ am Punkt $x$ ist:
+
+    \begin{center}
+        \eqboxf{$Hess_f (x) := \left(\dfrac{\partial^2 f_{(x)}}{\partial x_i \partial y_i}\right)_{1 \leq i,j \leq n} = \begin{pmatrix}
+                    \frac{\partial^2 f_{(x)}}{\partial x_1^2}            & \dots  & \frac{\partial^2 f_{(x)}}{\partial x_1 \partial x_n} \\
+                    \vdots                                               & \ddots & \vdots                                               \\
+                    \frac{\partial^2 f_{(x)}}{\partial x_n \partial x_1} & \dots  & \frac{\partial^2 f_{(x)}}{\partial x_n^2}            \\
+                \end{pmatrix}$}
+    \end{center}
+
+    $Hess_f (x)$ ist symmetrisch $\Rightarrow$ $Hess_f (x)$ \emph{diagonalisierbar} mit $\lambda_i \in \R$.
+
+    \subsubsection{Einschub: Definitheit einer Matrix}
+
+    \begin{tabular}{r l}
+        i)  & \hspace*{-13pt} Eine Matrix ist \textbf{positiv definit}, falls alle Eigenwerte $\lambda_i > 0$. \\
+        ii) & \hspace*{-13pt} Eine Matrix ist \textbf{negativ definit}, falls alle Eigenwerte $\lambda_i < 0$. \\
+        ii) & \hspace*{-13pt} Eine Matrix ist \textbf{indefinit}, falls positive und negative Eigenwerte.      \\
+    \end{tabular}
+
+    \subsubsection{Kritischer Punkt}
+
+    Ein Punkt $x_0 \in \Omega$ mit $df_{(x_0)} = 0$ (\emph{Koordinatenweise} $= 0$!) heisst kritischer Punkt von $f$.
+
+    \subsubsection{Satz}
+
+    Sei $f \in C^2(\Omega \subset \R^n)$. Ein kritischer Punkt $x_0$ ist eine
+
+    \begin{center}
+        \begin{tabular}{r l} \toprule
+            i)   & \hspace*{-10pt} strikte lokale Minimastelle, falls $Hess_f (x_0)$ positiv definit ist. \\
+            ii)  & \hspace*{-10pt} strikte lokale Maximastelle, falls $Hess_f (x_0)$ negativ definit ist. \\
+            iii) & \hspace*{-10pt} ein Sattelpunkt, falls $Hess_f (x_0)$ indefinit ist.                   \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    Bem: Bei degenerierten Punkten ($\det(Hess_f (x_0)) = 0$) kann man mit diesem Ansatz keine Aussage über lokales Min/Max/Sattelpunkt treffen!
+
+
+    \subsection{Vektorwertige Funktionen}
+
+    Sei $\Omega \subset \R^n$ offen und $f:\Omega \to \R^l$. $f$ heisst in $x_0 \in \Omega$ differenzierbar, falls alle Komponenten von $f$ in $x_0$ differenzierbar sind. \medskip
+
+    Das Differential mit Einheitsvektoren heisst \emph{Jacobi-Matrix} ($M_{l \times n} (\R)$):
+
+    \begin{center}
+        \eqbox{$df_{(x_0)} = \begin{pmatrix}
+                    df^1_{(x_0)} \\ \vdots \\ df^l_{(x_0)}
+                \end{pmatrix} =  \begin{pmatrix}
+                    \frac{\partial f^1}{\partial x_1} & \dots  & \frac{\partial f^1}{\partial x_n} \\
+                    \vdots                            & \ddots & \vdots                            \\
+                    \frac{\partial f^l}{\partial x_1} & \dots  & \frac{\partial f^l}{\partial x_n} \\
+                \end{pmatrix}$}
+    \end{center}
+
+    \subsubsection{Differentiationsregeln}
+
+    Sei $\Omega \subset \R^n$ offen. Seien $f,g: \Omega \to \R^l$ in $x_0 \in \Omega$ differenzierbar. Es gilt
+
+    \begin{center}
+        \begin{tabular}{r l} \toprule
+            i)  & \hspace*{-10pt} $d(f + g)_{x_0} = df_{(x_0)} + dg_{(x_0)}$                                                                                                \\
+            ii) & \hspace*{-10pt} $d \langle f, g \rangle_{(x_0)} = \sum\limits_{i = 1}^l \left( f^{i}(x_0) \cdot dg^{i}_{(x_0)} + g^{i}(x_0) \cdot df^{i}_{(x_0)} \right)$ \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    \subsubsection{Kettenregel 3te Version}
+
+    Seien $g: \R^n \to \R^l$ an $x_0 \in \R^n$ und $f: \R^l \to \R^m$ an $g(x_0)$ differenzierbar.
+
+    \begin{center}
+        \eqbox{$d(f \circ g)_{(x_0)} = \underbrace{df_{(g(x_0))}}_{\in M_{m \times l}(\R)} \cdot \underbrace{dg_{(x_0)}}_{\in M_{l \times n}(\R)}$}
+    \end{center}
+
+    Es empfiehlt sich stark für $g$ und $f$ andere Koordinaten zu benutzen!
+
+
+    \subsection{Der Umkehrsatz}
+
+    Sei $\Omega \subset \R^n$ offen und $f \in C^1(\Omega, \R^n)$. Sei $df_{(x_0)} \in M_{n \times n}(\R)$ invertierbar ($\det(df_{(x_0)}) \neq 0$) an einer Stelle $x_0 \in \Omega$. Dann existieren Umgebungen
+
+    \begin{center}
+        $\exists r > 0$ so dass $f:\underbrace{B_r(x_0)}_{:= U} \to \underbrace{f(B_r(x_0))}_{:=V}$
+    \end{center}
+
+    invertierbar ist und es existiert ein $g: V \to U$ so dass
+
+    \begin{center}
+        $g(f(x)) = x, \, \forall x \in U, \quad f(g(y)) = y, \, \forall y \in V$
+    \end{center}
+
+
+    und $g \in C^1(V, U)$ wobei $dg_{(f(x))} = (df_{(x)})^{-1}$. ($^{-1}$ $\hat{=}$ Matrixinverse!)
+
+
+    \subsubsection{Diffeomorphismus}
+
+    Sei $U,V \subset \R^n$ offen und $\Phi \in C^1(U;V)$. Die Abbildung $\Phi$ heisst ein \emph{Diffeomorphismus} von $U$ auf $\Phi(U) = V$, falls $\Phi$ \emph{injektiv ist} und falls die Umkehrabbildung $\Phi^{-1}$ von der Klasse $C^1(V;U)$ ist. \medskip
+
+    Aus dem Umkehrsatz folgt: Eine differenzierbare Abbildung mit invertierbarem Differential ist \emph{lokal} ein Diffeomorphismus.
+
+
+    \subsubsection{Anwendung: Polarkoordinanten}
+
+    Die Abbildung $f: (0, \infty) \times (-\pi,\pi)  \to \R^2$ mit
+
+    \begin{center}
+        $f(r, \varphi) = \begin{pmatrix}
+                r \cos(\varphi) \\ r \sin(\varphi)
+            \end{pmatrix}$ \qquad $df_{(r, \varphi)} = \begin{pmatrix}
+                \cos(\varphi) & - r \sin(\varphi) \\ \sin(\varphi) & r \cos(\varphi) \\
+            \end{pmatrix}$
+    \end{center}
+
+    erfüllt die Bedingungen vom Umkehrsatz, da
+
+    \begin{center}
+        $\det(df_{(r, \varphi)}) = r(\cos^2(\varphi) + \sin^2(\varphi)) = r > 0$
+    \end{center}
+
+    Man kann also \emph{lokal} folgende Umkehrabbildung einführen:
+
+    \begin{center}
+        $g:\begin{pmatrix}
+                x \\ y \\
+            \end{pmatrix} \mapsto \begin{pmatrix}
+                r = \sqrt{x^2 + y^2} \\ \varphi = \arctan(y/x) \\
+            \end{pmatrix}$
+    \end{center}
+
+
+    \subsection{Implizite Funktionen}
+
+    Das Ziel ist es die Levelmengen / Höhenlinien ($f^{-1}(\{c\})$) zu beschreiben.
+
+    \subsubsection{Satz}
+
+    Sei $\Omega \subset (\R^{n-1} \times \R)$ offen und $f \in C^1(\Omega, \R)$. Man hat also \emph{eine Variable der Funktion isoliert}: $f(\underbrace{x}_{\in \R^{n - 1}}, \underbrace{y}_{\in \R})$. \medskip
+
+    Sei ein Punkt $(x_0, y_0) \in \Omega$ mit $f(x_0, y_0) = 0$ und $\partial_y f_{(x_0,y_0)} \neq 0$. Dann existiert lokal eine Umgebung von $(x_0, y_0)$ und eine Funktion $h$:
+
+    \begin{center}
+        $h: B_r^{n - 1}(x_0) \to \R$
+    \end{center}
+
+    sodass, die Höhenlinie $f(x_0,y_0) = 0$ durch $h$ beschrieben wird:
+
+    \begin{center}
+        \eqbox{$\{(x,y) \in U; \, f(x,y) = 0\} \quad = \quad \{(x,h(x)); \, x \in B_r^{n-1}(x_0)\}$}
+    \end{center}
+
+    Insbesondere gilt $h(x_0) = y_0$ und $f(x_0,h(x_0)) = f(x_0, y_0)$. Ausserdem ist $h \in C^1(B_r^{n - 1}(x_0), \R)$ und das Differential ist:
+
+    \begin{center}
+        \eqbox{$\displaystyle dh_{(x_0)} = - \dfrac{1}{\partial_y f_{(x_0, h(x_0))}} \underbrace{\sum\limits_{i = 1}^{n - 1} \partial_{x_i} f_{(x_0, h(x_0))} \, dx_i}_{=d_x f}$}
+    \end{center}
+
+    wobei $d_x f$ das Differential von $f$ ist ohne den Koordinaten $y$.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Extrema mit Nebenbedingungen}
+
+    \subsubsection{Satz: Lagrange-Multiplikatorenregel}
+
+    Sei $f, g \in C^1(\R^n, \R)$ mit wiederum isolierter Variable: $(\underbrace{x}_{\in \R^{n - 1}}, \underbrace{y}_{\in \R})$.
+
+    Sei $(x_0, y_0)$ mit $g(x_0, y_0) = 0$ und $\partial_y g(x_0, y_0) \neq 0$. Und $(x_0, y_0)$ bildet \emph{ein lokales Minimum (bzw. Maximum)} für die Einschränkung von $f$ auf die Höhenlinie $g^{-1}(\{0\})$. Dann $\exists \lambda \in \R$ (Lagrange Multiplikator), so dass
+
+    \begin{center}
+        \eqboxf{$df_{(x_0,y_0)} + \lambda \cdot dg_{(x_0,y_0)} = 0$}
+    \end{center}
+
+    Bemerkung: Dies ist eine Addition von co-Vektoren, welches zu einem \textbf{Gleichungssystem} führt. (Koordinatenweise $= 0$!) \medskip
+
+    \textbf{Achtung}: Die Einschränkung $\partial_y g(x_0, y_0) \neq 0$ muss beachtet werden! Die Punkte, welche deswegen wegfallen können trotzdem ein Extrema sein, d.h. am Schluss vergleichen mit den Punkten vom Verfahren.
+
+    Beispiel: Auf $B_1^2(0)$ wären diese Punkte $(1,0)$ und $(-1,0)$.
+
+    \subsubsection{Nebenbedingungen: Einfache Randmengen}
+
+    Der Rand vom Einheitskreis ist: $\partial B_1(0) := \{(x,y) \in \R^2; x^2 + y^2 = 1\}$.
+
+    \subsection{Vorgehen: Globale Extremewerte bestimmen}
+
+    \begin{enumerate}
+        \item Argumentieren, wieso die Menge Kompakt ist.
+              \begin{itemize}
+                  \item Abgeschlossenheit: Menge duch \emph{stetige} Funktionen abgegrenzt, es folgt die Menge enthält alle Randpunkte.
+                  \item Beschränktheit: Auf die Ungleichung verweisen.
+                  \item Kompakt: Folgern das die Menge Kompakt ist und Extremumsatz gilt.
+              \end{itemize}
+        \item Kritische Punkte im Inneren bestimmen (Kandidaten für Extrema).
+        \item Alle Kandidaten für Extrema \emph{auf dem Rand der Menge} bestimmen. Man kann hier entweder Lagrange verwenden oder das alternative Vorgehen (siehe unten).
+        \item Die Randpunkte, welche nicht erfasst werden könnnen bestimmen, \textbf{auch Kandidaten}!
+        \item Alle Kandidaten in $f$ einsetzen und Minimum/Maximum bestimmen.
+    \end{enumerate}
+
+    \subsubsection{Alternatives Vorgehen für das Bestimmen der Kandidaten auf dem Rand}
+
+    Diese Vorgehen bietet sich gut an, wenn der Rand nicht durch eine einzige Nebenbedingungen darstellbar ist. Vorgehen:
+
+    \begin{enumerate}
+        \item Man parametrisiert den Rand mithilfe von Wegen $\gamma_i(t), \, t \in [a,b]$.
+        \item Man betrachtet für jeden Weg $\gamma_i$ die Funktion $f(\gamma_i)$ und analysiert, ob $f(\gamma_i)$ einen kritischen Punkt aufweist ($f'(\gamma_i) = 0$).
+        \item Man \textbf{überprüft}, ob der kritische Punkt überhaupt in $t$ drinnen ist.
+        \item Man nimmt zusätzlich \textbf{alle Randpunkte der Wege} als Kandidaten auf, d.h. $a, b$ eingesetzt in ihr $\gamma_i$ sind auch Kandidaten!
+    \end{enumerate}
+    \vfill\null
+    \pagebreak
+
+
+    \section{Wegintegrale}
+
+    \subsection{Differentialform (1-Form)}
+
+    Sei $\Omega \subset \R^n$ offen. Eine Abbildung $\lambda: \Omega \to L(\R^n,\R)$, welche jedem $x \in \Omega$ eine lineare Abbildung $\lambda(x): \R^n \to \R$ zuordnet, heisst Differentialform vom Grad 1. Es gilt ausserdem folgende Äquivalenz:
+
+    \begin{center}
+        \eqbox{1-Form: $\displaystyle \lambda(x) = \sum\limits_{i = 1}^n \lambda_i(x) d x^i$} $\Leftrightarrow$ Vektorfeld: $V(x) = \begin{pmatrix}
+                \lambda_1(x) \\ \vdots \\ \lambda_n(x)
+            \end{pmatrix}$
+    \end{center}
+
+    wobei $dx^{i} = (0, \dots, 0, 1, 0, \dots 0)$ die Basiselemente von $L(\R^n, \R)$ sind. \medskip
+
+    Bemerkung: Für jedes $f \in C^1(\Omega)$ ist das Differential $df$ eine 1-Form.
+
+
+    \subsection{Wegintegral}
+
+    Ein Weg ist eine vektorwertige Funktion $\gamma \in C^1_{stw}([a,b], \R^n)$. Man sagt:
+
+    \begin{center}
+        \begin{tabular}{r p{0.8\linewidth}} \toprule
+            i)  & \hspace*{-10pt} $\gamma(a)$ und $\gamma(b)$ sind seine Anfangs- und Endpunkte.         \\
+            ii) & \hspace*{-10pt} Die Ableitung $\dot \gamma$ ist der Geschwindigkeitsvektor.            \\
+            ii) & \hspace*{-10pt} Wenn $\gamma(a) = \gamma(b)$ gilt, dann heisst $\gamma$ abgeschlossen. \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    Sei $\gamma_1, \gamma_2 \in C^1_{stw}([0,1], \R^n)$ mit $\gamma_1(1) = \gamma_2(0)$. Dann ist $\gamma = \gamma_1 + \gamma_2$:
+
+    \begin{center}
+        $\gamma: [0,2] \to \Omega$ mit $t \mapsto \begin{cases}
+                \gamma_1(t)     & t \in [0,1] \\
+                \gamma_2(t - 1) & t \in [1,2] \\
+            \end{cases}$
+    \end{center}
+
+    Es gilt ausserdem: $\displaystyle \int_{\gamma_1 + \gamma_2} \lambda = \int_{\gamma_1} \lambda + \int_{\gamma_2} \lambda$
+
+    \subsubsection{Das Wegintegral}
+
+    Sei $\gamma \in C^1_{stw}([a,b], \R^n)$ und $\lambda$ eine 1-Form mit dem zu $\lambda$ zugehörigem Vektorfeld $V$. Dann ist das Wegintegral:
+
+    \begin{center}
+        \eqboxf{$\displaystyle \int_\gamma \lambda := \int\limits_a^b \lambda(\gamma(t)) \cdot \dot \gamma(t) dt \, \Leftrightarrow \,  \int_\gamma V \, d\vec{s} = \int\limits_a^b \langle V_i(\gamma(t)), \dot\gamma \rangle dt$}
+    \end{center}
+
+    Das Wegintegral ist \emph{unabhängig} von orientierungserhaltenden Umparametrisierungen von $\gamma$.
+
+
+    \subsubsection{Einschub: Wegzusammenhängend}
+
+    Sei $\Omega \subset \R^n$ offen. $\Omega$ heisst \emph{wegzusammenhängend}, falls jedes Paar von Punkten in $\Omega$ mit einem Weg verbunden werden kann.
+
+
+    \subsubsection{Satz}
+
+    Sei $\Omega$ offen und Wegzusammenhängend.
+
+    \begin{center}
+        \eqbox{$f \in C^1(\Omega)$ mit $df = 0 \, \Leftrightarrow f$ ist konstant}
+    \end{center}
+
+    \subsubsection{Einschub: Parametrisierungen}
+
+    \begin{tabular}{r l l}
+        Gerade von $a$ nach $b$:        & \hspace*{-12pt} $\gamma(t) = (1-t) \cdot a + t \cdot b$                       & \hspace*{-5pt} $t \in [0,1]$     \\
+        Kreis (positiven Sinne):        & \hspace*{-12pt} $\gamma(t) = (r\cdot \cos(\varphi), r \cdot \sin(\varphi))$   & \hspace*{-5pt} $t \in [0, 2\pi]$ \\
+        Kreis (\emph{negativen} Sinne): & \hspace*{-12pt} $\gamma(t) = (r\cdot \cos(\varphi), - r \cdot \sin(\varphi))$ & \hspace*{-5pt} $t \in [0, 2\pi]$ \\
+        Ellipse (positive Sinne):       & \hspace*{-12pt} $\gamma(t) = (a \cdot \cos(t), b \cdot \sin(t))$              & \hspace*{-5pt} $t \in [0,2\pi]$  \\
+    \end{tabular}
+
+
+
+
+    \subsection{Potentiale}
+
+    Sei $\Omega \subset \R^d$ offen und $\lambda \in C^0(\Omega, \R^n)$ eine 1-Form. Es sind \textbf{äquivalent}: \medskip
+
+    i) $\exists f \in C^1(\Omega)$ mit \eqbox{$\lambda = df$} \quad ($f$ heisst ''Potential von $\lambda$'') \medskip
+
+    ii) Für je zwei Wege $\gamma_1, \gamma_2 \in C^1_{stw}([a,b], \Omega)$ mit $\gamma_1(a) = \gamma_2(a), \, \gamma_1(b) = \gamma_2(b)$, d.h. mit \emph{gleichen Anfangs- und Endpunkten}, gilt:
+
+    \begin{center}
+        \eqboxf{$\displaystyle \int\limits_{\gamma_1} \lambda = \int\limits_{\gamma_2} \lambda = f(\gamma(b)) - f(\gamma(a))$}
+    \end{center}
+
+    iii) Für jeden \textbf{geschlossenen} Weg $\gamma \in C^1_{stw}([a,b], \Omega)$ mit $\gamma(a) = \gamma(b)$:
+
+    \begin{center}
+        \eqbox{$\displaystyle \int\limits_{\gamma} \lambda = 0$} \quad (''$V$ ist \emph{konservativ}'')
+    \end{center}
+
+    Bem: Potentiale sind bis auf die Addition einer Konstante bestimmt!
+
+
+    \subsubsection{Verfahren zur Berechnung eines Potentials}
+
+    Sei $\lambda$ eine 1-Form. Das zugehörige Potential $f$, falls es existiert, kann man mit folgendem Verfahren ermitteln: \medskip
+
+    i) Wir setzen oBdA:  $f(0,0,0) = 0$ \medskip
+
+    ii) Wir nehmen die folgenden drei Wegintegrale:
+
+    \begin{center}
+        $\gamma_1(t) = \begin{pmatrix}
+                t \, x \\ 0 \\ 0 \\
+            \end{pmatrix}, \, \gamma_2(t) = \begin{pmatrix}
+                x \\ t \, y \\ 0 \\
+            \end{pmatrix}, \, \gamma_3(t) = \begin{pmatrix}
+                x \\ y \\ t \, z \\
+            \end{pmatrix}$
+    \end{center}
+
+    iii) Aus dem zweiten Punkt im vorherigen Satz folgt:
+
+    \begin{center}
+        $\displaystyle f(x,y,z) = f(0,0,0) + \int_{\gamma_1 + \gamma_2 + \gamma_3} \lambda = \int_{\gamma_1} \lambda + \int_{\gamma_2} \lambda + \int_{\gamma_3} \lambda$
+    \end{center}
+
+    \begin{center}
+        \begin{minipage}{0.35\linewidth}
+            \begin{center}
+                \includegraphics[width=1\linewidth]{Bilder/Potential.png}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.64\linewidth}
+            iv) Zum Schluss: \textbf{Verifizieren}, dass $f$ wirklich das Potential von $\lambda$ ist:
+
+            \begin{center}
+                \eqbox{$df(x,y,z) = \lambda$}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \subsubsection{Satz: Potentialfeld}
+
+    Für ein Vektorfeld $V \in C^0(\Omega, \R^n)$ sind äquivalent:
+
+    \begin{center}
+        \eqbox{$V$ ist konservativ $\Leftrightarrow$ $\exists f \in C^1(\Omega): V = \nabla f$}
+    \end{center}
+
+    In diesem Fall heisst $V$ \textbf{Potentialfeld} mit dem Potential $f$.
+
+
+    \subsubsection{Korollar: Rotationsvektorfeld}
+
+    Sei $V = (V_1, \dots, V_n) \in C^1(\Omega, \R^n)$ konservativ. Dann gilt
+
+    \begin{center}
+        \eqbox{$\dfrac{\partial V^{i}}{\partial x^{j}} - \dfrac{\partial V^{j}}{\partial x^{i}} = 0, \quad \forall \, 1 \leq i,j \leq n$}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \section{Integration in $\R^n$}
+
+
+    \subsection{Riemannsches Integral über einen Quader}
+
+    Ein $n$-dimensionaler \emph{Quader} ist ein Produkt von Intervallen
+
+    \begin{center}
+        $\displaystyle Q = \prod\limits_{i = 1}^n I_i = \{x = (x^{i})_{1 \leq i \leq n}; x^{i} \in I_i, 1 \leq i \leq n\}$
+    \end{center}
+
+    Solch ein Quader $Q$ hat den folgenden \emph{Elementarinhalt}:
+
+    \begin{center}
+        \eqbox{$\displaystyle \mu(Q) = \mu([a,b] \times [c,d]) = \prod\limits_{i = 1}^n \left| I_i \right|$}
+    \end{center}
+
+    \begin{center}
+        \begin{minipage}{0.74\linewidth}
+            Die \emph{Zerlegung} $P = \{Q_k; 1 \leq k \leq K\}$ eines Quaders in disjunkte Teilquader $Q = \bigcup\limits_{k = 1}^K$ hat folgende \emph{Feinheit}:
+
+            \begin{center}
+                $\delta_P = \max\limits_{1 \leq k \leq K} diam (Q_k)$
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.25\linewidth}
+            \begin{center}
+                \includegraphics[width=1\linewidth]{Bilder/Quader.png}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    wobei $diam (Q_k)$ den \emph{Durchmesser} von $Q_k$ bezeichnet:
+
+    \begin{center}
+        $diam(Q_k) = \sup\limits_{x,y \in Q_k} \left| x - y \right|, \, 1 \leq k \leq K$
+    \end{center}
+
+    \subsubsection{Treppenfunktion in $\R^n$}
+
+    Eine Funktion $f: Q \to \R$ auf einem Quader $Q$ heisst Treppenfunktion, falls $f(x)$ eine Darstellung folgender Form besitzt
+
+    \begin{center}
+        \eqbox{$\displaystyle f(x) = \sum\limits_{k = 1}^K c_k \cdot \chi_{Q_k}(x)$ wobei $\chi_{Q_k}(x) = \begin{cases}
+                    1 & , x \in Q_k \\ 0 & , x \not\in Q_k \\
+                \end{cases}$}
+    \end{center}
+
+    mit einer Zerlegung $P = \{Q_k; 1 \leq k \leq K\}$ und Konstanten $c_k \in \R$. \medskip
+
+    Das Riemann-Integral einer Treppenfunktion $f(x)$ ist, wie folgt definiert:
+
+    \begin{center}
+        \eqbox{$\displaystyle\int_Q f(x) \,d\mu = \sum\limits_{k = 1}^K c_k \cdot \mu(Q_k)$}
+    \end{center}
+
+    \subsubsection{Satz: Verfeinerung der Zerlegung}
+
+    Eine Zerlegung $\tilde{P} = \{\tilde{Q}_j; \, 1 \leq j \leq J\}$ ist eine Verfeinerung der Zerlegung $P = \{Q_k; 1 \leq k \leq K\}$, falls jedes $\tilde{Q}_j$ in einem Quader $Q_k$ enthalten ist. \medskip
+
+    Das Integral wird durch eine Verfeinerung \textbf{nicht} verändert.
+
+
+    \subsection{Das Riemann Integral}
+
+    Seien $e^-,e^+$ Treppenfunktionen und $f: Q \to \R$ beschränkt. Dann gilt:
+
+    \begin{center}
+        Untere R-Integral von $f$: \, \eqbox{$\displaystyle\underline{\int_Q} f(x) \,d\mu = \sup\limits_{e^-(x) \leq f(x)} \int_Q e^-(x) \,d\mu$} \medskip
+
+        Obere R-Integral von $f$: \eqbox{$\displaystyle\overline{\int_Q} f(x) \,d\mu = \inf\limits_{f(x) \leq e^+(x)} \int_Q e^+(x) \,d\mu$}
+    \end{center}
+
+    Die Funktion $f$ heisst \emph{R-integrabel} über $Q$, falls
+
+    \begin{center}
+        \eqboxf{$\displaystyle\underline{\int_Q} f(x) \,d\mu = \overline{\int_Q} f(x) \,d\mu =: \int_Q f(x) \,d\mu$}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsubsection{Satz}
+
+    Sei $f \in C^0(Q)$. Dann ist $f$ über $Q$ R-integrabel.
+
+    \subsubsection{Satz: Riemannsche Summen}
+
+    Für jede Folge von Zerlegungen $(P^{(l)})_{l \in \N}$ von $Q$ mit Feinheit $\delta_{P^{(l)}} \xrightarrow[]{l \to \infty} 0$ gilt für eine beliebige Auswahl von Punkten $x_k^{(l)} \in Q_k^{(l)}$ stets
+
+    \begin{center}
+        \eqbox{$\displaystyle \int_Q \left(\sum\limits_{k = 1}^{K^{(l)}} f(x_k^{(l)}) \chi_{Q_k^{(l)}}\right) d \mu = \sum\limits_{k = 1}^{K^{(l)}} f(x_k^{(l)}) \cdot \mu(Q_k^{(l)}) \xrightarrow[]{l \to \infty} \int_Q f \,d\mu$}
+    \end{center}
+
+
+    \subsection{Eigenschaften des Riemannschen Integrals}
+
+    \subsubsection{Linearität}
+
+    Seien $f, g: Q \to \R$ beschränkt und über $Q$ R-integrabel, und $a \in \R$.
+
+    \begin{center}
+        \eqbox{$\displaystyle \int_Q \left( \alpha \cdot f(x) + g(x) \right) \,d\mu = \alpha \cdot \int_Q f(x) \,d\mu + \int_Q  g(x) \,dx$}
+    \end{center}
+
+    \subsubsection{Monotonie}
+
+    Seien $f,g: Q \to \R$ beschränkt und über $Q$ R-integrabel, und sei $f \leq g$.
+
+    \begin{center}
+        \eqbox{$\displaystyle \int_Q f(x) \, d\mu \leq \int_Q g(x) \, d\mu$}
+    \end{center}
+
+    Insbesondere gilt für $f \in C^0(\overline{Q})$ die Abschätzung
+
+    \begin{center}
+        \eqbox{$\displaystyle \left| \int_Q f(x) \, d\mu \right| \leq \int_Q \left| f(x) \right| \, d\mu \leq \sup\limits_{Q} \left| f(x) \right| \cdot \mu(Q)$}
+    \end{center}
+
+    \subsubsection{Korollar}
+
+    Seien $f, f_k \in C^0(\overline{Q})$ mit $f_k \xrightarrow[]{glm.} f(k \to \infty)$. Dann gilt
+
+    \begin{center}
+        \eqbox{$\displaystyle\left| \int_Q f_k \,d\mu - \int_Q f \,d\mu \right| \leq \int_Q | f_k - f | \, d\mu \leq ||f_k - f || \cdot \mu(Q) \xrightarrow[]{(k \to \infty)} 0$}
+    \end{center}
+
+    \subsubsection{Gebietsadditivität}
+
+    Sei $f: Q \to \R$ beschränkt und über $Q$ R-integrabel. Sei $P = \{Q_k; \, 1 \leq k \leq K\}$ eine Zerlegung von $Q$ in disjunkte Quader $Q_k$. Dann gilt
+
+    \begin{center}
+        \eqbox{$\displaystyle\int_Q f(x) \,d\mu = \sum\limits_{k = 1}^K \int_{Q_k} f(x) \,d\mu$}
+    \end{center}
+
+
+
+    \subsection{Satz von Fubini}
+
+    Sei $Q = [a,b] \times [c,d] \subset \R^2$ \textbf{und} sei $f \in C^0(Q)$. Dann gilt
+
+    \begin{center}
+        \eqboxf{$\displaystyle\int_Q f(x,y) \,d\mu = \int\limits_a^b \left( \int\limits_c^d f(x,y) \,dy \right) dx = \int\limits_c^d \left( \int\limits_a^b f(x,y) \,dx \right) dy$}
+    \end{center}
+
+    Analoges gilt auch in höheren Dimensionen, solange $f \in C^0(Q)$ ist.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Jordan-Bereiche}
+
+    Sei $\Omega \subset \R^n$ beschränkt und $Q$ ein \emph{beliebiger} Quader mit $\Omega \subset Q$.
+
+    \begin{center}
+        \begin{minipage}{0.25\linewidth}
+            \begin{center}
+                \includegraphics[width=1\linewidth]{Bilder/Jordan-Messbar.png}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.74\linewidth}
+            Sei $\chi_\Omega$ die charakteristische Funktion:
+
+            \begin{center}
+                $\chi_\Omega(x) = \begin{cases}
+                        1 & x \in \Omega \\ 0 & x \not\in \Omega \\
+                    \end{cases}$
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \subsubsection{Jordan-messbar (JM)}
+
+    Die Menge $\Omega$ heisst \emph{Jordan-messbar}, falls $\chi_\Omega(x)$ R-integrabel über $Q$ ist, diese Eigenschaft ist vom Quader $Q$ das $\Omega$ enthält \textbf{unabhängig}. \medskip
+
+    Das n-dimensionale Jordansche Mass von $\Omega$: \eqbox{$\displaystyle \mu(\Omega) = \int_Q \chi_{\Omega}(x) \,d\mu$}
+
+    \subsubsection{Satz}
+
+    Seien $\Omega_1, \Omega_2$ \emph{JM}. Dann sind $\Omega_1 \cap \Omega_2$ und $\Omega_1 \cup \Omega_2$ auch \emph{JM} und
+
+    \begin{center}
+        \eqbox{$\mu(\Omega_1) + \mu(\Omega_2) = \mu(\Omega_1 \cup \Omega_2) + \mu(\Omega_1 \cap \Omega_2)$}
+    \end{center}
+
+    \subsubsection{R-Integral über Jordan-messbare Bereiche}
+
+    Sei $\Omega \subset Q$ \emph{JM} und $f: \Omega \to \R$ beschränkt. $f$ heisst \textbf{R-Integrabel über} $\Omega$ , falls die Fortsetzung $\overline{f}$ von $f$ über $Q$ R-integrabel ist und es gilt:
+
+    \begin{center}
+        \eqbox{$\overline{f} = \begin{cases}
+                    f(x) & x \in \Omega \\ 0 & x \in Q \setminus \Omega \\
+                \end{cases} \qquad \displaystyle \int_\Omega f\, d\mu := \int_Q \overline{f}\, d\mu$}
+    \end{center}
+
+    \subsubsection{Satz}
+
+    Sei $\Omega$ \emph{JM} und $f \in C^0(\Omega)$ beschränkt. Dann ist $f$ auf $\Omega$ R-Integrabel.
+
+
+    \subsection{Hypograph und Hypergraph}
+
+    Sei $Q' \subset \R^{n-1}$ und $\psi \in C^0_{stw}(\overline{Q'})$ mit $\psi \geq 0$. Dann ist die Menge
+
+    \begin{center}
+        \eqbox{$\Omega_{\psi} = \{ (x', x_n) \in \R^n ; \, x' \in Q', \, 0 \leq x_n \leq \psi(x') \}$}
+    \end{center}
+
+    Jordan-messbar und heisst \textbf{Hypograph}. \medskip
+
+    Sei $Q' \subset \R^{n-1}$ und $\phi  \in C^0_{stw}(\overline{Q'})$ mit $\phi \leq 0$. Dann ist die Menge
+
+    \begin{center}
+        \eqbox{$\Omega_{\phi} = \{ (x', x_n) \in \R^n ; \, x' \in Q', \, \phi(x') \leq x_n \leq 0 \}$}
+    \end{center}
+
+    Jordan-messbar und heisst \textbf{Hypergraph}. \medskip
+
+    Die Beschränkung der Menge ist flexibel, also man kann auch:
+
+    \begin{center}
+        \eqboxf{$\Omega = \{ (x, y) \in \R^2 ; \, x \in [a,b] , \,\phi(x) \leq y \leq \psi(x)\}$}
+    \end{center}
+
+    Das zugehörige Integral ist für den Fall $\R^2$:
+
+    \begin{center}
+        \eqbox{$\displaystyle \int_{\Omega} f \, d\mu = \int\limits_{a}^b \left(\, \int\limits_{\phi(x)}^{\psi(x)} f(x,y) dy \right) dx$}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Satz von Green}
+
+    Sei $\Omega \subset Q \subset \R^2$ ein Gebiet der Klasse $C^1_{stw}$ und $g, h \in C^1(\overline{\Omega})$. Es gilt
+
+    \begin{center}
+        \eqboxf{$\displaystyle \int_\Omega \left(\dfrac{\partial h}{\partial x} (x,y) - \dfrac{\partial g}{\partial y}(x,y) \right) \,d\mu = \int_{\partial\Omega} \underbrace{g(x,y) \,dx + h(x,y)\, dy}_{:= \lambda}$}
+    \end{center}
+
+    wobei der Rand von $\Omega$ so parametrisiert wird, dass $\Omega$ \textbf{zur Linken} liegt.
+
+    \subsubsection{Gebiet der Klasse $C^1_{stw}$ in $\R^2$}
+
+    Sei $\Omega \subset Q$ (beschränkt). Ein Gebiet $\Omega$ ist von der Klasse $C^1_{stw}$, falls \textbf{zu jedem} Punkt $p \in \partial\Omega$ folgendes existiert:
+
+    \begin{center}
+        \begin{minipage}{0.74\linewidth}
+            i) Falls notwendig, \emph{eine Drehung der Koordinatenachsen}, d.h. von $(x,y)$ zu $(x_1, x_2)$:
+
+            \begin{center}
+                $\begin{pmatrix}
+                        x_1 \\ x_2 \\
+                    \end{pmatrix} = \begin{pmatrix}
+                        \cos(\Theta)  & \sin(\Theta) \\
+                        -\sin(\Theta) & \cos(\Theta) \\
+                    \end{pmatrix} \begin{pmatrix}
+                        x \\ y \\
+                    \end{pmatrix}$
+            \end{center}
+
+            ii) Es $\exists$ ein Quader $[a,b] \times [c,d]$ in den neuen Koordinaten $(x_1, x_2)$, welcher $p \in (a,b) \times (c,d)$ erfüllt.
+        \end{minipage}
+        \begin{minipage}{0.25\linewidth}
+            \begin{center}
+                \includegraphics[width=1\linewidth]{Bilder/Stueckweise_Gebit.png}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    iii) Es $\exists \psi \in C^1_{stw}([a,b],[c,d])$ und die Schnittmenge $\Omega \cap (a,b)\times(c,d)$ ist ein \emph{Hyper-/Hypograph}.
+
+    \subsubsection{Satz von Green mit Vektorfeld}
+
+    Sei $\Omega \subset \R^2$ beschränkt und von der Klasse $C^1_{stw}$. Sei $V = \begin{pmatrix}
+            V_1(x,y) \\ V_2(x,y) \\
+        \end{pmatrix}$ ein $C^1(\overline{\Omega};\R^2)$ Vektorfeld. Dann gilt:
+
+
+    \begin{center}
+        \begin{minipage}{0.54\linewidth}
+            \begin{center}
+                \eqbox{$\text{rot(V)} = \partial_x V_2 (x,y) - \partial_y V_1(x,y)$}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.45\linewidth}
+            \begin{center}
+                \eqboxf{$\displaystyle \int_\Omega \text{rot(V)} \,d\mu = \int_{\partial\Omega} V \, d\vec{s}$}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    wobei $\text{rot}(V)$ die Rotation von $V$ ist und $\partial\Omega$ ist so parametrisiert, dass $\Omega$ \textbf{zur Linken} liegt. \medskip
+
+    Bem: Wenn $rot(V) = 1$ ist, dann kann man \emph{die Fläche} $\mu(\Omega)$ berechnen.
+
+
+    \subsection{Satz von Poincaré}
+
+    Sei $\Omega \subset \R^2$ ein beschränktes, einfach zusammenhängendes $C^1_{stw}$ Gebiet und sei $V$ ein $\R^2$-Vektorfeld der Klasse $C^1(\overline{\Omega}, \R^2)$. Dann gilt:
+
+    \begin{center}
+        \eqboxf{$V$ ist konservativ ($V = \nabla f$) $\Leftrightarrow$ rot($V) = 0$}
+    \end{center}
+
+    \subsubsection{Einschub: Einfach zusammenhängende Gebiete}
+
+    Sei $\Omega \subset \R^2$ beschränkt, von der Klasse $C^1_{stw}$ und wegzusammenhängend. $\Omega$ heisst einfach zusammenhängend, falls $\partial\Omega$ nur eine 'Komponente' hat. \medskip
+
+    Informell: $\Omega$ besitzt keine 'Löcher'.
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Substitutionsregel}
+
+    \subsubsection{Einschub: Diffeomorphismus}
+
+    Sei $U,V \subset \R^n$ offen und $\Phi \in C^1(U;V)$. Die Abbildung $\Phi$ heisst ein Diffeomorphismus von $U$ auf $\Phi(U) = V$, falls $\Phi$ \emph{injektiv ist} und falls die Umkehrabbildung $\Phi^{-1} \in C^1(V;U)$. \medskip
+
+    Aus dem Umkehrsatz folgt: $\Phi \in C^1(U;V)$ Diffeomorphismus $\Leftrightarrow$
+
+    \begin{center}
+        \eqbox{ $\Phi$ injektiv und $det(d\Phi_{(x_0)}) \neq 0, \forall x_0 \in U$}
+    \end{center}
+
+    \subsubsection{Transformationssatz}
+
+    Sei $U,V \subset \R^n$ offen und $\Phi \in C^1(U,V)$ ein Diffeomorphismus. Sei $\overline{\Omega} \subset U$ beschränkt und Jordan messbar. Dann ist $\Phi(\Omega)$ Jordan messbar, und
+
+    \begin{center}
+        \eqbox{$\displaystyle \mu(\Phi(\Omega)) = \int_{\Omega} |\det(d\Phi(x))| d\mu(x)$}
+    \end{center}
+
+    wobei $\mu(\Phi(\Omega))$ das \textbf{Volumen} von $\Phi(\Omega)$ ist.
+
+    \subsubsection{Satz: Substitutionsregel}
+
+    Sei $U,V \subset \R^n$ offen, $\Phi \in C^1(U,\R^n)$ ein Diffeomorphismus von $U$ auf $V \subset \R^n$. Sei $\Omega \subset U$ beschränkt und Jordan messbar, und sei $f:\Phi(\Omega) \to \R$ beschränkt und R-integrabel. Dann gilt
+
+    \begin{center}
+        \begin{minipage}{0.6\linewidth}
+            \begin{center}
+                \eqboxf{$\displaystyle \int_{\Phi(\Omega)} f \, d\mu = \int_{\Omega} (f \circ \Phi) \cdot |\det(d\Phi)| \, d\mu$}
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.39\linewidth}
+            \begin{center}
+                \includegraphics[width=1\linewidth]{Bilder/Substitutionsregel.png}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \subsubsection{Einschub: Verschiede Koordinatentransformationen}
+
+    Die folgenden Koordinatentransformationen sind ein Diffeomorphismus, wie von der Substitutionsregel verlangt. \medskip
+
+
+    \textbf{Polarkoordinaten}: $\Phi: (0, \infty) \times (-\pi,\pi) \to \R^2$ mit
+
+    \begin{center}
+        $\Phi(r, \varphi) := \begin{pmatrix}
+                r \cos(\varphi) \\ r \sin(\varphi)
+            \end{pmatrix}$ \quad und \quad $\det(d\Phi_{(r, \varphi)}) = r$
+    \end{center}
+
+    Desweiteren gilt: Bild($\Phi) = \R^2 \setminus \{(x,y) \in \R^2; \, y = 0, x \leq 0\}$. \medskip
+
+    \textbf{Zylinderkoordinaten}: $\Phi: (0, \infty) \times (-\pi,\pi) \times \R \to \R^3$ mit
+
+    \begin{center}
+        $\Phi(r, \varphi, h) := \begin{pmatrix}
+                r \cos(\varphi) \\ r \sin(\varphi) \\ h \\
+            \end{pmatrix}$ \quad und \quad $\det(d\Phi_{(r, \varphi,h)}) = r$
+    \end{center}
+
+    Desweiteren gilt: Bild($\Phi) = \R^3 \setminus \{(x,y,z) \in \R^3; \, y = 0, x \leq 0\}$. \medskip
+
+    \textbf{Kugelkoordinaten}: $\Phi: (0, \infty) \times (0,\pi) \times (-\pi,\pi) \to \R^3$ mit
+
+    \begin{center}
+        $\Phi(r, \Theta, \varphi) := \begin{pmatrix}
+                r \sin(\Theta) \cos(\varphi) \\ r \sin(\Theta) \sin(\varphi) \\ r \cos(\Theta) \\
+            \end{pmatrix}$ \quad und \quad $\det(d\Phi_{(r, \Theta, \varphi)}) = r^2 \sin(\Theta)$
+    \end{center}
+
+    Desweiteren gilt: Bild($\Phi) = \R^3 \setminus \{(x,y,z) \in \R^3; \, y = 0, x \leq 0\}$.
+    \vfill\null
+    \columnbreak
+
+
+
+
+    \subsection{Oberflächenmass und Flussintegral}
+    %https://video.ethz.ch/lectures/d-math/2022/spring/401-0232-00L/25701150-98ca-4c82-a6fe-f812ac4074a9.html
+
+    \subsubsection{Lokale Immersion}
+
+    Sei $U \subset \R^2$ offen und $\Phi \in C^1(U, \R^3)$ \emph{injektiv}. $\Phi$ bildet eine lokale Immersion, falls $d\Phi(x) \in M_{3 \times 2}(\R)$ \textbf{für alle} $(x,y) \in U$ regulär ist ($Rang = 2$).
+
+    \begin{center}
+        \eqbox{Rang$(d\Phi) = 2 \Leftrightarrow \dfrac{\partial \Phi}{\partial x}$, $\dfrac{\partial \Phi}{\partial y}$ linear unabhängig $\Leftrightarrow \dfrac{\partial \Phi}{\partial x} \times \dfrac{\partial \Phi}{\partial y} \neq 0$}
+    \end{center}
+
+    Bem: Diese Bedingung ist notwendig, damit es lokal eine Ebene in $\R^3$ abbildet (schöne Oberfläche), sonst wäre es lokal teils eine Linie. \medskip
+
+    Bem: Sei $\psi \in C^1(Q, \R)$ wobei $Q = [a,b] \times [c,d]$ mit $\phi(x,y) := \begin{pmatrix}
+            x \\ y \\ \psi(x,y) \\
+        \end{pmatrix}$. Dann ist $\phi(x,y)$ \emph{immer} eine lokale Immersion.
+
+    \subsubsection{Der Oberflächeninhalt}
+
+    Sei $\Phi: U \subset \R^2 \to \R^3$ eine lokale Immersion. Sei $\overline{\Omega} \subset U$ beschränkt und Jordan messbar. Der 2-dimensionale Flächeninhalt von $S = \Phi(\overline{\Omega})$ ist
+
+    \begin{center}
+        \eqboxf{$\displaystyle \mu_2(\Phi(\Omega)) := \int_\Omega \underbrace{|| \partial_x \Phi \times \partial_y \Phi|| d\mu}_{=: d o} = \int_S do$}
+    \end{center}
+
+    wobei $do$ der \emph{skalare Flächeninhalt} bezüglich $\Phi$ ist.
+
+
+    \subsubsection{Das Integral einer Funktion über eine Oberfläche}
+
+    Sei $\Phi: U \subset \R^2 \to \R^3$ eine lokale Immersion. Sei $\overline{\Omega} \subset U$ beschränkt und Jordan messbar. Sei $S = \Phi(\Omega)$ das zugehörige Flächenstück in $\R^3$ und $f: \overline{S} \to \R$ stetig. Dann ist das Integral von $f$ auf $S$:
+
+    \begin{center}
+        \eqbox{$\displaystyle \int_S f do := \int_\Omega (f \circ \Phi) \cdot ||\partial_x \Phi \times \partial_y \Phi || d \mu$}
+    \end{center}
+
+
+    \subsubsection{Normalenvektor}
+
+    Sei $\Phi: U \subset \R^2 \to \R^3$ eine lokale Immersion. Der Normaleinheitsvektor $n$ zur Fläche $\Phi(U) = S$ ist:
+
+    \begin{center}
+        \eqbox{$\vec{n} = \dfrac{\partial_x \Phi \times \partial_y \Phi}{|| \partial_x \Phi \times \partial_y \Phi ||}$}
+    \end{center}
+
+    \subsubsection{Das Flussintegral}
+
+    Sei $V = \begin{pmatrix}
+            P \\ Q \\ R \\
+        \end{pmatrix}$ ein stetiges Vektorfeld. Sei $\Phi: U \subset \R^2 \to \R^3$ eine lokale Immersion. Sei $\overline{\Omega} \subset U$ beschränkt und Jordan messbar. Dann ist der Fluss von $V$ durch die Fläche $S = \Phi(\overline{\Omega})$:
+
+    \begin{center}
+        \eqboxf{$\displaystyle \int_S V \cdot \vec{n} \, do = \int_\Omega (V \circ \Phi) \cdot \dfrac{\partial_x \Phi \times \partial_y \Phi}{|| \partial_x \Phi \times \partial_y \Phi ||} || \partial_x \Phi \times \partial_y \Phi || d \mu$}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Der Satz von Stokes in $\R^3$}
+    %https://video.ethz.ch/lectures/d-math/2022/spring/401-0232-00L/25701150-98ca-4c82-a6fe-f812ac4074a9.html @45min
+
+    \subsubsection{Die Rotation eines $\R^3$ Vektorfeld}
+
+    Sei ein Vektorfeld $V \in C^1(\R^3)$. Dann ist die \emph{Rotation} von $V$:
+
+    \begin{center}
+        $\text{rot}(V) = \nabla \times V = \begin{pmatrix}
+                \partial_{x_1} \\ \partial_{x_2} \\ \partial_{x_3} \\
+            \end{pmatrix} \times \begin{pmatrix}
+                V_1 \\ V_2 \\ V_3 \\
+            \end{pmatrix} = \begin{pmatrix}
+                \partial_{x_2}V_3 - \partial_{x_3}V_2 \\ \partial_{x_3}V_1 - \partial_{x_1}V_3 \\ \partial_{x_1}V_2 - \partial_{x_2}V_1 \\
+            \end{pmatrix}$
+    \end{center}
+
+
+    \subsubsection{Satz von Stokes}
+
+    Sei $\Phi$ eine lokale Immersion von $U \subset \R^2 \to \R^3$. Sei $\overline{\Omega} \subset U$ in $C^1_{stw}$, Jordan messbar und beschränkt. Sei $V \in C^1(\R^3)$. Dann gilt:
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.25}
+        \eqbox{\begin{tabular}{r l}
+                $\displaystyle \int_{S = \Phi(\Omega)} \text{rot}(V) \cdot \vec{n} \,d o$ & $\displaystyle= \int_\Omega (\text{rot}(V) \circ \Phi) \cdot (\partial_x \Phi \times \partial_y \Phi) d\mu$                   \\
+                                                                                          & $\displaystyle= \int_{\partial\Omega} (V \circ \Phi) \cdot \frac{d}{dt}(\Phi \circ \gamma) dt = \int_{\partial S} V d\vec{s}$ \\
+            \end{tabular}}
+    \end{center}
+
+    In Worten: Der Fluss von $\text{rot}(V)$ durch eine Fläche gleicht der Zirkulation vom Vektorfeld $V$ seinem Rand entlang.
+
+    \begin{center}
+        \includegraphics[width=0.75\linewidth]{Bilder/Stokes.png}
+    \end{center}
+
+
+    \subsection{Der Satz von Gauss}
+
+    \subsubsection{Divergenz eines Vektorfeldes}
+
+    Die Divergenz (''Quellstärke'') eines Vektorfeldes $V \in C^1(\R^3)$ ist:
+
+    \begin{center}
+        $\text{div}(V) = \partial_x V_1(x,y,z) + \partial_y V_2 (x,y,z) + \partial_z V_3 (x,y,z)$
+    \end{center}
+
+    \subsubsection{Satz von Gauss in $\R^2$}
+
+    Sei $\Omega$ ein $C^1_{stw}$ Gebiet mit dem zu $\partial\Omega$ zugehörigen Tangentialvektor $\dot\gamma$. Sei $\nu$ der Normalvektor zum Rand (''äussere Normale''). Dann gilt:
+
+    \begin{center}
+        \begin{minipage}{0.4\linewidth}
+            \begin{center}
+                $\nu  = \begin{pmatrix} \dot \gamma_2 \\ - \dot\gamma_1 \\
+                    \end{pmatrix} \dfrac{1}{\sqrt{\dot\gamma_1^2 + \dot\gamma_2^2}}$
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.59\linewidth}
+            \begin{center}
+                \eqbox{$\displaystyle \int_\Omega \text{div}(V) = \int_{\partial\Omega} V \cdot \nu \, d\vec{s}$}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    \subsubsection{Satz von Gauss in $\R^3$}
+
+    Sei ein Vektorfeld $V \in C^1(\R^3)$. Sei $\psi \in C^1(Q := [a,b] \times [c,d], \R_+)$ mit dem Hypographen $\Omega_{\psi} = \{x \in \R^3; (x_1,x_2) \in Q, 0  \leq x_3 \leq \psi(x_1,x_2)\}$. Wir setzen vorraus, dass $V = 0$ für $x_3 \leq 0$ und für $(x_1,x_2) \not\in Q$. Es gilt:
+
+    \begin{center}
+        \eqbox{$\displaystyle \int_{\Omega_\psi} \text{div}(V) d\mu = \int_Q V(\Phi) \cdot (\partial_x \Phi \times \partial_y \Phi)\, d\mu$} $\phi(x,y) := \begin{pmatrix}
+                x \\ y \\ \psi(x,y) \\
+            \end{pmatrix}$
+    \end{center}
+
+    In Worten: Das Integral von $\text{div}(V)$ in $\Omega_{\psi}$ gleicht dem Fluss von $V$ durch die Fläche von $\psi(x,y)$.
+
+
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Beispiel eines Oberflächenintegrals}
+
+    Wir berechnen die Oberfläche $S = \Phi(B_1^2(0))/2$ (nur obere Halbkugel).
+
+    \begin{center}
+        \begin{minipage}{0.6\linewidth}
+            \begin{center}
+                $\Phi(x,y) = \begin{pmatrix}
+                        x \\ y \\ \sqrt{1 - x^2 - y^2} \\
+                    \end{pmatrix}$
+            \end{center}
+        \end{minipage}
+        \begin{minipage}{0.39\linewidth}
+            \begin{center}
+                \includegraphics[width=0.4\linewidth]{Bilder/Oberflachenintegral_BSP.png}
+            \end{center}
+        \end{minipage}
+    \end{center}
+
+    Wir bekommen also:
+
+    \begin{center}
+        $\partial_x \Phi = \begin{pmatrix}
+                1 \\ 0 \\ - \frac{x}{\sqrt{1 - x^2 - y^2}} \\
+            \end{pmatrix} \qquad \partial_y \Phi = \begin{pmatrix}
+                0 \\ 1 \\ - \frac{y}{\sqrt{1 - x^2 - y^2}} \\
+            \end{pmatrix}$ \medskip
+
+        $\partial_x \Phi \times \partial_y \Phi = \begin{pmatrix}
+                \frac{x}{\sqrt{1 - x^2 - y^2}} \\ \frac{y}{\sqrt{1 - x^2 - y^2}} \\ 1 \\
+            \end{pmatrix} \Rightarrow || \partial_x \Phi \times \partial_y \Phi || = \dfrac{1}{\sqrt{1 - (x^2 + y^2)}}$
+    \end{center}
+
+    Wir benutzen nun die Substitutionsregel mit Polarkoordinaten:
+
+    \begin{center}
+        $\mu_2(S) = \displaystyle \int\limits_{B_1^2(0)} || \partial_x \Phi \times \partial_y \Phi|| d\mu = \int\limits_{-\pi}^\pi \left( \int\limits_0^1 \dfrac{r}{\sqrt{1 - r^2}} dr \right) d\Theta = 2\pi$
+    \end{center}
+
+    Wir integrieren nun die Höhe $f(x,y,z) = z$ auf der Oberfläche $S$:
+
+    \begin{center}
+        $\displaystyle \int\limits_S  f do = \int\limits_{-\pi}^{\pi} \left( \int\limits_0^1 \dfrac{r \cdot \sqrt{1 - r^2}}{\sqrt{1 - r^2}} dr \right) d\Theta = 2\pi \left[\dfrac{r^2}{2}\right]^1_0 = \pi = \dfrac{1}{2} \mu_2(S)$
+    \end{center}
+
+
+    \subsection{Punktmengen}
+
+    Sei der Radius $r > 0$ und der Index 0 markiert das Zentrum. Dann gilt: \medskip
+
+    Kreis: $K = \left\{(x, y)\in \mathbb{R}^2; (x-x_0)^2+(y-y_0)^2 = r^2 \right\}$ \medskip
+
+    Kugel: $K = \left\{(x, y, z)\in \mathbb{R}^3; (x-x_0)^2+(y-y_0)^2+(z-z_0)^2 = r^2 \right\}$ \medskip
+
+    Zylinder: $Z=\left\{(x, y, z)\in \mathbb{R}^3; (x-x_0)^2+(y-y_0)^2 =r^2, \, 0\le z \le h \right\}$ \medskip
+
+    Kegel: $K=\left\{(x, y, z)\in \mathbb{R}^3; x^2+y^2 = \frac{r^2}{h^2}(h-z)^2 \right\}$ \medskip
+
+    Ellipse: $E=\left\{(x, y)\in \mathbb{R}^2; \frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2 }{b^2} = r^2 \right\}$ \medskip
+
+    Ellipsoid: $E=\left\{(x, y, z)\in \mathbb{R}^3; \frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2} +\frac{(z-z_0)^2}{c^2} = r^2 \right\}$
+
+    \subsubsection{Volumen eines Ellipsoid}
+
+    Für eine Berechnung des Volumens eines Ellipsoids benutzt man die folgendermassen angepassten Kugelkoordinanten:
+
+    \begin{center}
+        $F(r, \Theta, \varphi) := \begin{pmatrix}
+                a \cdot r \sin(\Theta) \cos(\varphi) \\ b \cdot r \sin(\Theta) \sin(\varphi) \\ c \cdot r \cos(\Theta) \\
+            \end{pmatrix}$ und $\det(dF_{(r, \Theta, \varphi)}) = abc \cdot r^2 \sin(\Theta)$
+    \end{center}
+
+    $\int_{E(a, b, c, R)} 1 d \mu=\int_{B_{R}(0)}|\det d F(x, y, z)| d z d y d x = abc \mu\left(B_{R}(0)\right)$
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Kochrezepte}
+
+
+    \subsubsection{Integralgrenzen von einem Hyper- und Hypograph bestimmen}
+
+    \begin{enumerate}
+        \item Die Variable für das äusserste Integral wählen (oft $x$), alle anderen Variablen in der Mengengleichung auf $0$ setzen. Nun kann man die Grenze für die erste Variable herauslesen.
+        \item Die Variable vom zweitäussersten Integral (oft $y$) auswählen, alle anderen Variablen in der Menge \textbf{ausser die schon bestimmte Variable} auf $0$ setzen.
+        \item[2.5] Analoger Schritt für die 3te Variable. 
+        \item Nun hat man die Integralgrenzen bestimmt und kann fortfahren mit der Berechnung vom Integral.
+    \end{enumerate}
+
+
+    \subsubsection{Kochrezept Volumenberechnung}
+
+    \begin{enumerate}
+        \item Das Integrationsgebiet $\Phi(\Omega)$ (bzw. Integrationsgrenzen) bestimmen in den passenden Koordinatentransformationen.
+
+              \textbf{Achtung}: Auf die Einschränkungen der Koordinatentransformation achten (z.B. $r \in ]0,\pi[$).
+        \item[1.5] Evtl. bemerken, dass die Koordinatentransformation eine Halbebene nicht trifft, dies aber vernachlässigbar ist beim Transformationssatz.
+        \item Den Transformationssatz anwenden ($\det(d\Phi)$ nicht vergessen).
+    \end{enumerate}
+
+
+    \subsubsection{Kochrezept Oberflächeninhalt}
+
+    \begin{enumerate}
+        \item Die Menge anschauen und bestimmen um was für ein Objekt es sich handelt, Skizzen helfen! \textbf{Man berechnet die Oberfläche stückweise}. Einfache Fläche, wie Kreisflächen, kann man direkt mit den bekannten Formeln berechnen.  
+        \item Schwerere Oberflächen muss man mit folgendem Vorgehen berechnen: 
+        \begin{itemize}
+            \item Eine Achse vorläufig entfernen, d.h. Variable auf $0$ setzen in der Mengengleichung. Man schaut von nun an von dieser Achse aus auf das Objekt. (z.B.: Man entfernt $z$ $\Rightarrow$ man schaut von oben).
+            \item Skizze von dem neuen 2D-Gebiet erstellen und dann die Ungleichungen (am Ende die Integralgrenzen) der zwei übrig bleibenden Variablen für das 2D-Gebiet bestimmen. 
+            \item Mit der ursprünglichen Mengengleichung eine Funktion $\psi(x,y)$ für die entfernte Variable bestimmen. \textbf{Achtung}: Es kann sein, dass hier zwei Funktionen herauskommen, in diesem Fall muss man den Oberflächeninhalt für \textbf{beide} berechnen (evtl. Symmetrie!). 
+            \item Lokale Immersion der Form $\phi(x,y) := \begin{pmatrix}
+                x \\ y \\ \psi(x,y) \\
+            \end{pmatrix}$ bilden.
+            \item Oberflächeninhalt(e) berechnen mit der bekannten Formel, das Integrationsgebiet ist das zuvor bestimmte 2D-Gebiet. 
+        \end{itemize} 
+        \item \textbf{Alle} Oberflächeninhalt zusammenaddieren. 
+    \end{enumerate}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Einfache Geometrieformeln}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{r l l} \toprule
+                      & Fläche/Volumen              & Umfang/Oberfläche                      \\
+            \midrule
+            Kreis     & $A = \pi r^2$               & $U = 2 \pi r$                          \\
+            Kugel     & $V = \frac{4}{3} \pi r^3$   & $S = 4 \pi r^2$                        \\
+            Ellipsoid & $V = \frac{4}{3} \pi a b c$ &                                        \\
+            Zylinder  & $V = \pi r^2 h $            & $S = 2\pi r h + 2\pi r^2$              \\
+            Kegel     & $V = \frac{1}{3} \pi r^2 h$ & $S = \pi r^2 + \pi r \sqrt{h^2 + r^2}$ \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+\end{multicols*}
+
+\input{sections/Ubersicht.tex}
+
+
+\setcounter{secnumdepth}{2}
+\end{document}
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+
+\begin{multicols*}{3}
+    \subsection{Tabelle mit Ableitungen und Stammfunktionen}
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.75}
+        \begin{tabular} {r c c c l} \toprule
+            $f'(x)$                               & \hspace*{-10pt}
+            $\xleftrightharpoons[\int f(x) dx]{\frac{d}{dx}}$
+            \hspace*{-10pt}                       & $f(x)$          & \hspace*{-10pt}
+            $\xleftrightharpoons[\int f(x) dx]{\frac{d}{dx}}$
+            \hspace*{-10pt}                       & $F(x)$                                                                                \\
+            \midrule
+            $n \cdot x^{n - 1}$                   & \hspace*{-20pt} & $x^n$              & \hspace*{-20pt} & $\dfrac{1}{n + 1} x^{n + 1}$ \\
+            $e^x$                                 & \hspace*{-20pt} & $e^x$              & \hspace*{-20pt} & $e^x$                        \\
+            $\dfrac{1}{x}$                        & \hspace*{-20pt} & $\log|x|$          & \hspace*{-20pt} & $x (\log|x| - 1)$            \\
+            \midrule
+            $\cos(x)$                             & \hspace*{-20pt} & $\sin(x)$          & \hspace*{-20pt} & $-\cos(x)$                   \\
+            $-\sin(x)$                            & \hspace*{-20pt} & $\cos(x)$          & \hspace*{-20pt} & $\sin(x)$                    \\
+            $\frac{1}{\cos^2(x)} = 1 + \tan^2(x)$ & \hspace*{-20pt} & $\tan(x)$          & \hspace*{-20pt} & $-\log|\cos(x)|$             \\
+            \midrule
+            $\dfrac{1}{\log(a) \cdot x}$          & \hspace*{-20pt} & $\log_a|x|$        & \hspace*{-20pt} &                              \\
+            $\log(a) \cdot a^x$                   & \hspace*{-20pt} & $a^x$              & \hspace*{-20pt} & $\dfrac{1}{\log(a)} a^x$     \\
+            $x^x (\log(x)+1)$                     & \hspace*{-20pt} & $x^x$              & \hspace*{-20pt} &                              \\
+            \midrule
+            $\cosh(x)$                            & \hspace*{-20pt} & $\sinh(x)$         & \hspace*{-20pt} &                              \\
+            $\sinh(x)$                            & \hspace*{-20pt} & $\cosh(x)$         & \hspace*{-20pt} &                              \\
+            $\dfrac{1}{\cosh^2(x)}$               & \hspace*{-20pt} & $\tanh(x)$         & \hspace*{-20pt} & $\log(\cosh(x))$             \\
+            \midrule
+            $\dfrac{1}{\sqrt{1 - x^2}}$           & \hspace*{-20pt} & $\arcsin(x)$       & \hspace*{-20pt} &                              \\
+            $- \dfrac{1}{\sqrt{1 - x^2}}$         & \hspace*{-20pt} & $\arccos(x)$       & \hspace*{-20pt} &                              \\
+            $\dfrac{1}{1 + x^2}$                  & \hspace*{-20pt} & $\arctan(x)$       & \hspace*{-20pt} &                              \\
+            \midrule
+            $\dfrac{1}{\sqrt[]{x^2 + 1}}$         & \hspace*{-20pt} & $\text{arsinh}(x)$ & \hspace*{-20pt} &                              \\
+            $\dfrac{1}{\sqrt[]{x^2 - 1}}$         & \hspace*{-20pt} & $\text{arcosh}(x)$ & \hspace*{-20pt} &                              \\
+            $\dfrac{1}{1 - x^2}$                  & \hspace*{-20pt} & $\text{artanh}(x)$ & \hspace*{-20pt} &                              \\
+            \midrule
+            $\text{sign}(x) = \begin{cases}
+                                      -1 & x < 0 \\  1 & 0 < x \\
+                                  \end{cases}$ \hspace*{-10pt}       & \hspace*{-20pt} & $|x|$              & \hspace*{-20pt} &               \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    Bemerkung: Bei Ableitungen mit Logarithmen, sowie den inversen Trigo- und Hyperfunktionen ist der Definitionsbereich eingeschränkt!
+
+
+    \subsection{Stetige Funktionen}
+
+    Folgende Elementarfunktionen sind stetig auf ihrem Definitionsbereich:
+
+    \begin{center}
+        \begin{tabular}{l} \toprule
+            i) Polynome sind stetige Funktionen auf $\R$.                                                 \\
+            ii) Rationale Funktionen $\frac{p}{q}$ sind stetig auf $\Omega = \{ z \in \C; q(z) \neq 0\}$. \\
+            iii) Die Wurzelfunktion ist auf $\R_+$ stetig.                                                \\
+            iv) Die Exponentialfunktion ist auf $\R$ stetig.                                              \\
+            v) Die Logartihmusfunktion ist auf $]0, \infty[$ stetig.                                      \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+
+    \subsection{Partialbruchzerlegung}
+
+    Ziel: Rationale Funktionen $\left(\frac{p(x)}{q(x)}\right)$ in Teilbrüche zerlegen. Vorgehen: \medskip
+
+    i) Wenn der Zähler einen höheren Grad als den Nenner hat, muss man zuerst eine Polynomdivision durchführen, d.h. $\left(p(x):q(x) = \dots\right)$. \medskip
+
+    ii) Das Nennerpolynom $q(x)$ in Nullstellenform bringen.
+
+    \begin{center}
+        $q(x) = \prod\limits_{i = 1} (x - x_i)^{m_i}$ \qquad wobei $(x - x_i) = 0$
+    \end{center}
+
+    iii) a) Die Nullstelle hat Multiplizität 1 ($m_i = 1$):
+
+    \begin{center}
+        $\dfrac{C}{(x - x_i)}$
+    \end{center}
+
+    b) Die Nullstelle hat Multiplizität grösser 1 ($1 < m_i$):
+
+    \begin{center}
+        $\dfrac{C_1}{(x - x_i)} + \dfrac{C_2}{(x - x_i)^2} + \dots + \dfrac{C_{m_i}}{(x - x_i)^{m_i}}$
+    \end{center}
+
+    c) Die Nullstelle ist komplexwertig (Gewünschte Form: $ax^2 + b x + c$):
+
+    \begin{center}
+        $\dfrac{Ax + B}{(a x^2 + b x + c)}$
+    \end{center}
+
+    iii) Alle Koeffizienten $C_i$ bestimmen durch einen Koeffizientenvergleich.
+    \vfill\null
+    \columnbreak
+
+
+    \section{Ergänzungen aus LinAlg}
+
+    \subsection{Determinante}
+
+    Sei $A \in M_{2 \times 2}(\R)$. Dann ist die Determinante:
+
+    \begin{center}
+        \eqbox{$\det\begin{bmatrix}
+                    a & b \\ c & d \\
+                \end{bmatrix} := a d - b c$}
+    \end{center}
+
+
+    \subsubsection{Laplace Entwicklung}
+
+    Sei die Matrix $A = \begin{bmatrix}
+            a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\
+        \end{bmatrix}$. Die Entwicklung nach Zeile 2 ist:
+
+    \begin{center}
+        $\det(A) = -a_2 \cdot \det\begin{bmatrix}
+                b_1 & c_1 \\ b_3 & c_3 \\
+            \end{bmatrix} + b_2 \cdot \det\begin{bmatrix}
+                a_1 & c_1 \\ a_3 & c_3 \\
+            \end{bmatrix} - c_2 \cdot \det \begin{bmatrix}
+                a_1 & b_1 \\ a_3 & b_3 \\
+            \end{bmatrix}$
+    \end{center}
+
+    Für die Vorzeichen gilt zu beachten: $\begin{bmatrix}
+            + & - & + & - \\
+            - & + & - & + \\
+            + & - & + & - \\
+            - & + & - & + \\
+        \end{bmatrix}$
+
+
+    \subsection{Eigenwerte und Eigenvektoren}
+
+    Die Eigenwerte einer Matrix $A$ berechnet man mit
+
+    \begin{center}
+        \eqbox{$\det(A - \lambda I) = 0$}
+    \end{center}
+
+    Der zum Eigenwert $\lambda_i$ dazugehörige Eigenvektor $\vect{s}_i$ berechnet man durch das Auflösen von folgendem homogenen LGS:
+
+    \begin{center}
+        \eqbox{$(A - \lambda_i I) \cdot \vect{s}_i = \vect{0}$}
+    \end{center}
+
+
+    \subsubsection{Diagonalisierbar}
+
+    Sei $A \in M_{n \times n}(\R)$ mit $n$ linear unabhängigen Eigenvektoren $\vect{s}_1$ und seien $\lambda_1, \dots, \lambda_n$ die Eigenwerte von $A$. Dann ist $A$ diagonalisierbar:
+
+    \begin{center}
+        \eqbox{$A = S D S^{-1} = [\vect{s}_1 \dots \vect{s}_n] \begin{bmatrix}
+                \lambda_1 &        & 0         \\
+                          & \ddots &           \\
+                0         &        & \lambda_n \\
+            \end{bmatrix} [\vect{s}_1 \dots \vect{s}_n]^{-1}$}
+    \end{center}
+
+
+    \subsection{Matrixinverse berechen}
+
+    Zuerst Gauss-Elimination, dann Rücksubstitution ($[A | I] \Rightarrow [I | A^{-1}]$)
+
+    \subsubsection{Explizite Formeln}
+
+    Für $A \in M_{2 \times 2}(\R)$ gilt:
+
+    \begin{center}
+        \eqbox{$A^{-1} = \begin{bmatrix}
+                a & b \\
+                c & d \\
+            \end{bmatrix}^{-1} = \dfrac{1}{\det(A)} \cdot
+            \begin{bmatrix}
+                d  & -b \\
+                -c & a  \\
+            \end{bmatrix}$}
+    \end{center}
+
+    Für $A \in M_{3 \times 3}(\R)$ gilt:
+
+    \begin{center}
+        $A^{-1} = \begin{bmatrix}
+                a & b & c \\
+                d & e & f \\
+                g & h & i \\
+            \end{bmatrix}^{-1} = \dfrac{1}{\det(A)} \cdot
+            \begin{bmatrix}
+                ei - fh & ch - bi & bf - ce \\
+                fg - di & ai - cg & cd - af \\
+                dh - eg & bg - ah & ae - bd \\
+            \end{bmatrix}$
+    \end{center}
+\end{multicols*}
+
+
+\begin{multicols*}{2}
+    \section{Spass mit Integralen}
+
+    \subsection{Tangenssubstitution}
+
+    Sei $t(x) = \tan(\frac{x}{2})$ mit $x \in ]-\pi, \pi[$. Dann gilt
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{l l} \toprule
+            $\cos(x) = \dfrac{1 - t^2(x)}{1 + t^2(x)}$ & $\sin(x) = \dfrac{2 t(x)}{1 + t^2(x)}$   \\
+            $\cos^2(x) = \dfrac{1}{1 + t^2(x)}$        & $\sin^2(x) = \dfrac{t^2(x)}{1 + t^2(x)}$ \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+    Mit dieser Substitution kann man gewisse Trigonometrische Integrale einfacher lösen.
+    \subsection{Rückwärtssubstitution}
+
+    Die Substitutionsregel lässt sich auch rückwärts durchführen. Sei $\varphi(x)$ \emph{injektiv}. Dann gilt:
+
+    \begin{center}
+        \eqboxf{$\displaystyle \int\limits_{\alpha}^{\beta} f(t) dt = \int\limits_{\varphi^{-1}(\alpha)}^{\varphi^{-1}(\beta)} f\left( \varphi(x) \right) \cdot \varphi'(x) dx$}
+    \end{center}
+
+    Bei geschickter Wahl der Funktion $\varphi(x)$ kann entgegen des ersten Anscheins der Integrand vereinfacht werden.
+
+    \subsubsection{Tabelle}
+    %https://de.wikipedia.org/wiki/Integration_durch_Substitution#Anwendung
+    %https://en.wikipedia.org/wiki/Trigonometric_substitution
+
+    Bem: Nach Anwendung der Regel ist die Trigo-Identiät ($cos^2(x) + sin^2(x) = 1$) notwendig!
+
+    \begin{center}
+        \renewcommand{\arraystretch}{2}
+        \begin{tabular}{c l l l} \toprule
+            \textbf{Integral}                                             & \multicolumn{2}{l}{\textbf{Rücksubstitution}}                                                                                          \\
+            \midrule 
+            $\int\limits_{\alpha}^{\beta} \sqrt{1 - t^2} dt$              & $\varphi(x) = \sin(x)$                        & $\varphi^{-1}(t) = \arcsin(t)$                       & $\varphi'(x) = \cos(x)$         \\
+            \midrule
+            $\int\limits_{\alpha}^{\beta} \sqrt{a^2 - t^2} dt$            & $\varphi(x) = a \cdot \sin(x)$                & $\varphi^{-1}(t) = \arcsin\left(\dfrac{t}{a}\right)$ & $\varphi'(x) = a \cdot \cos(x)$ \\
+            $\int\limits_{\alpha}^{\beta} \dfrac{1}{\sqrt{a^2 - t^2}} dt$ & $\varphi(x) = a \cdot \sin(x)$                & $\varphi^{-1}(t) = \arcsin\left(\dfrac{t}{a}\right)$ & $\varphi'(x) = a \cdot \cos(x)$ \\
+            \midrule
+            $\int\limits_{\alpha}^{\beta} \sqrt{1 + t^2} dt$              & $\varphi(x) = \sinh(x)$                       & $\varphi^{-1}(t) = \text{arsinh}(t)$                 & $\varphi'(x) = \cosh(x)$        \\
+            $\int\limits_{\alpha}^{\beta} \sqrt{t^2 - 1} dt$              & $\varphi(x) = \cosh(x)$                       & $\varphi^{-1}(t) = \text{arcosh}(t)$                 & $\varphi'(x) = \sinh(x)$        \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+
+
+    \subsection{Integrale über eine Periode (Orthogonalitätsrelationen)}
+
+    Sei $\omega = \dfrac{2\pi}{T}$ und $m,n \in \N$. Dann gelten folgende Relationen:
+
+    \begin{center}
+        \begin{tabular}{l c l} \toprule
+            $\displaystyle\int\limits_0^T \sin(n \omega t) dt = 0$                             & \hspace*{+20pt} &
+            $\displaystyle\int\limits_0^T \cos(n \omega t) dt = 0$                                                                                                                                                         \\
+            \midrule
+            $\displaystyle\int\limits_0^T \sin(n \omega t) \sin(m \omega t) dt = \begin{cases}
+                                                                                         0 & n \neq m \\ \frac{T}{2} & n = m \\
+                                                                                     \end{cases}$ & \hspace*{+20pt} & $\displaystyle\int\limits_0^T \cos(n \omega t) \cos(m \omega t) dt = \begin{cases}
+                                                                                                                                                                                               0 & n \neq m \\ \frac{T}{2} & n = m \\
+                                                                                                                                                                                           \end{cases}$ \\
+            \midrule
+            $\displaystyle\int\limits_0^T \sin(n \omega t) \cos(m \omega t) dt = 0$                                                                                                                                        \\
+            \toprule
+        \end{tabular}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+    \subsection{Liste von Trigonometrischen Integralen}
+
+    Man kann diese Integrale \emph{normalerweise} benutzen bei der Prüfung, solange man auf die Identität vermerkt. Man setzt dabei einfach die Integralgrenzen ein, wie man es intuitiv machen würde.
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{l c l} \toprule
+            $\displaystyle \int \sin^2(x) dx = \dfrac{x - \sin(x)\cos(x)}{2} + C$                                    & \hspace*{+10pt} & $\displaystyle \int\frac{1}{\sin{(x)}}dx =\ln{\vert\frac{\sin{(x)}}{\cos{(x)}+1}\vert} + C$             \\
+            $\displaystyle \int \cos^2(x) dx = \dfrac{x + \sin(x)\cos(x)}{2} + C$                                    & \hspace*{+10pt} & $\displaystyle \int\frac{1}{\cos{(x)}}dx =\ln{\vert\frac{-\cos{(x)}}{\sin{(x)}-1}\vert} + C$            \\
+            $\displaystyle \int \sin(x) \cos(x) dx = \dfrac{sin^2(x)}{2} + C$                                        & \hspace*{+10pt} & $\displaystyle \int \frac{1}{\tan(x)} dx =\ln\vert \sin(x) \vert + C$                                   \\
+            $\displaystyle \int \sin^2(x)\cos(x)dx = \frac{1}{3}\sin^3(x) + C$                                       & \hspace*{+10pt} & $\displaystyle \int \frac{1}{\cos^2(x)}dx =\tan(x) + C$                                                 \\
+            $\displaystyle \int \sin(x)\cos^2(x)dx = -\frac{1}{3}\cos^3(x) + C$                                      & \hspace*{+10pt} & $\displaystyle \int \frac{1}{\sin^2{(x)}}dx =-\frac{1}{\tan{(x)}} + C$                                  \\
+            $\displaystyle \int \sin^2(x)\cos^2(x)dx = \frac{1}{32}(4x-\sin(4x)) + C$                                & \hspace*{+10pt} & $\displaystyle \int \arcsin(x)dx = x\cdot \arcsin(x)+\sqrt{1-x^2} + C$                                  \\
+            $\displaystyle \int \arccos(x)dx =x\cdot \arccos(x)-\sqrt{1-x^2} + C$                                    & \hspace*{+10pt} & $\displaystyle \int \arctan(x)dx =x\cdot \arctan(x)-\frac{1}{2}\ln \vert x^2+1\vert + C$                \\
+            \midrule
+            $\displaystyle \int_0^{2\pi}\cos^4(t)\text{dt} =\displaystyle \int_0^{2\pi}\sin^4(t)dt = \frac{3\pi}{4}$ & \hspace*{+10pt} & $\displaystyle \int_0^{2\pi}\cos^3(t)\text{dt} =\displaystyle \int_0^{2\pi}\sin^3(t)dt = 0$             \\
+            $\displaystyle \int_0^{2\pi}\cos^2(t)\text{dt} =\displaystyle \int_0^{2\pi}\sin^2(t)dt = \pi$            & \hspace*{+10pt} & $\displaystyle \int_0^{2\pi}\sin(t)\cos^2(t)\text{dt} =\displaystyle \int_0^{2\pi}\cos(t)\sin^2(t)dt=0$ \\
+            $\displaystyle \int_0^{2\pi}\sin(t)\cos(t)\text{dt} =0$                                                  & \hspace*{+10pt} &                                                                                                         \\
+            \toprule
+        \end{tabular}
+    \end{center}
+
+    \subsubsection{Tabelle von ausgewerteten Integralen}
+
+    Mit der Begründung ''Symmetrie'' ist es normalerweise erlaubt die \emph{Nullintegrale} der Tabelle zu benutzen. \medskip
+
+    Den Rest der Tabelle würde ich nur zur Überprüfung der Resultate an der Prüfung verwenden. Denke nicht, dass es Pünkte gibt, wenn man direkt das Resultat schreibt.
+
+    \begin{center}
+        \renewcommand{\arraystretch}{1.5}
+        \begin{tabular}{ r c c c c c c c }\toprule
+            \textbf{Funktion:}     & \multicolumn{5}{l}{\textbf{Integralgrenzen:}}                                                                                                                                                                                                                                                       \\
+            \midrule
+                                   & $\displaystyle\int_0^{\frac{\pi}{4}}$         & $\displaystyle\int_0^{\frac{\pi}{2}}$ & $\displaystyle\int_0^{\pi}$ & $\displaystyle\int_0^{2\pi}$ & $\displaystyle\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$ & $\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$ & $\displaystyle\int_{-\pi}^{\pi}$ \\
+            \midrule
+            $\sin(x)$              & $\frac{\sqrt{2}-1}{\sqrt{2}}$                 & $1$                                   & $2$                         & $0$                          & $0$                                                  & $0$                                                  & $0$                              \\
+
+            $\sin^2(x)$            & $\frac{\pi-2}{8}$                             & $\frac{\pi}{4}$                       & $\frac{\pi}{2}$             & $\pi$                        & $\frac{\pi-2}{4}$                                    & $\frac{\pi}{2}$                                      & $\pi$                            \\
+
+            $\sin^3(x)$            & $\frac{8-5\sqrt{2}}{12}$                      & $\frac{2}{3}$                         & $\frac{4}{3}$               & $0$                          & $0$                                                  & $0$                                                  & $0$                              \\
+
+            $\cos(x)$              & $\frac{1}{\sqrt{2}}$                          & $1$                                   & $0$                         & $0$                          & $\sqrt{2}$                                           & $2$                                                  & $0$                              \\
+
+            $\cos^2(x)$            & $\frac{2+\pi}{8}$                             & $\frac{\pi}{4}$                       & $\frac{\pi}{2}$             & $\pi$                        & $\frac{2+\pi}{4}$                                    & $\frac{\pi}{2}$                                      & $\pi$                            \\
+
+            $\cos^3(x)$            & $\frac{5}{6\sqrt{2}}$                         & $\frac{2}{3}$                         & $0$                         & $0$                          & $\frac{5}{3\sqrt{2}}$                                & $\frac{4}{3}$                                        & $0$                              \\
+
+            $\sin \cdot \cos(x)$   & $\frac{1}{4}$                                 & $\frac{1}{2}$                         & $0$                         & $0$                          & $0$                                                  & $0$                                                  & $0$                              \\
+
+            $\sin^2 \cdot \cos(x)$ & $\frac{1}{6\sqrt{2}}$                         & $\frac{1}{3}$                         & $0$                         & $0$                          & $\frac{1}{3\sqrt{2}}$                                & $\frac{2}{3}$                                        & $0$                              \\
+
+            $\sin \cdot \cos^2(x)$ & $\frac{4-\sqrt{2}}{12}$                       & $\frac{1}{3}$                         & $\frac{2}{3}$               & $0$                          & $0$                                                  & $0$                                                  & $0$                              \\
+            \bottomrule
+        \end{tabular}
+    \end{center}
+\end{multicols*}
+
+\begin{multicols*}{3}
+    \section{Relevante Plots}
+
+    \subsection{Trigonometrische Funktionen}
+
+    \begin{center}
+        \includegraphics[width=1\linewidth]{Bilder/Trigonometric_functions.png}
+    \end{center}
+
+    \subsection{Einheitskreis}
+
+    \begin{center}
+        \includegraphics[width=1\linewidth]{Bilder/unit-circle.jpg}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+    \subsection{Hyperbelfunktionen}
+
+    \begin{center}
+        \includegraphics[width=1\linewidth]{Bilder/Sinh_cosh_tanh.png}
+    \end{center}
+    \vfill\null
+    \columnbreak
+
+    \subsection{Areafunktionen (Umkehrfunktionen der Hyperbelfunktionen)}
+
+    \begin{center}
+        \includegraphics[width=0.95\linewidth]{Bilder/arsinh.png}
+    \end{center}
+
+    \begin{center}
+        \includegraphics[width=0.95\linewidth]{Bilder/arcosh.png}
+    \end{center}
+
+    \begin{center}
+        \includegraphics[width=0.95\linewidth]{Bilder/artanh.png}
+    \end{center}
+\end{multicols*}
+
+\begin{multicols*}{3}
+    \subsection{Kochrezepte}
+
+    \subsubsection{Überprüfung auf Differenzierbarkeit}
+
+    Meistens ist der Ursprung $(0,0)$ gefragt mit Funktionen, welche bis auf den Ursprung differenzierbar sind. Das allgemeine Vorgehen ist: \medskip
+
+    i) Auf Stetigkeit überprüfen. \textbf{Polarkoordinantentrick}:
+
+    \begin{center}
+        $r^2 = x^2+y^2$ mit $x = r \cdot \cos(\varphi)$ und $y = r \cdot \sin(\varphi)$
+    \end{center}
+    \medskip
+
+    Falls $\lim\limits_{r \to r_0}$ unabhängig von $\varphi$ existiert, dann ist $f$ stetig in $(x_0,y_0)$. \medskip
+
+    \emph{Unstetigkeit zeigen}: Man untersucht die Grenzwerte verschiedener Folgen $(\frac{1}{n},\frac{1}{n})$ und $(0,\frac{1}{n})$ und zeigt, dass zwei Unterschiedliche Grenzwerte vorhanden sind. \medskip
+
+    ii) Differenzierbarkeit überprüfen: Partielle Ableitungen bestimmen und Definition Differenzierbarkeit einsetzen (evtl. Polarkoordinantentrick für Grenzwert benutzen). \medskip
+
+    \emph{Nicht differenzierbar} zeigen: Neben Unstetigkeit in $(x_0,y_0)$ oder Unstetigkeit von $\partial_x f, \partial_y f$ kann man auch zeigen, dass für $\vec{v} = h\cdot(v_1,v_2)$:
+
+    \begin{center}
+        $\lim\limits_{h\to 0} \dfrac{f(h v_1,h v_2)-f(x_0)}{h}$
+    \end{center}
+
+    unterschiedliche Werte, z.B. links- und rechtsseitiger Grenzwert sind nicht gleich, besitzt. Oder man zeigt, dass die Richtungsableitungen nicht linear von $v$ abhängen.
+
+
+    \subsubsection{Überprüfen auf Stetigkeit}
+
+    Neben den schon oben erwähnten Tricks, gibt es noch ein Paar weitere Hinweise: \medskip
+
+    Beim $\delta,\epsilon$-Kriterium oder gleichmässige Konvergenz benötigt man oft die Dreiecksungleichung (oft mit einer verschwindenden $\pm$-Term Addition) oder die binomischen Formeln.
+
+
+    \subsubsection{Überprüfung Gleichmässige Konvergenz}
+
+    \begin{enumerate}
+        \item Punktweisen Limes von $f_n$ auf $\Omega$ für fixes $x\in\Omega$ berechnen, d.h.
+              $$f(x) =\lim \limits_{n\to\infty} f_n(x)$$
+
+              Kann verschiedene Werte annehmen, je nach Punkt $x_0$!
+
+        \item Prüfe $f_n$ auf gleichmässige Konvergenz. Vorgehen:
+              \begin{enumerate}
+                  \item Berechne $\sup \limits_{x\in\Omega} |f_n(x)-f(x)|$. Oft ist es von Vorteil die \textbf{Ableitung} $\frac{d}{dx} |f_n(x)-f(x)|$ zu berechnen und \textbf{gleich Null zu setzen}, um das Maximum der Menge zu bestimmen.
+                  \item Bilde den Limes für $n\to\infty$: $\lim\limits_{n\to\infty}\sup\limits_{x\in\Omega}|f_n(x)-f(x)|$, konvergiert dies für $n\to \infty$ so gilt gleichmässige konvergenz.
+              \end{enumerate}
+              Indirekte Methoden:
+              \begin{enumerate}
+                  \item $ f$ unstetig $\Rightarrow$ keine gleichmässige Konvergenz
+                  \item $f$ stetig, $f_n(x)\leq f_{n+1}(x),\quad\forall x \in\Omega $ und $\Omega$ kompakt $\Rightarrow$ gleichmässige Konvergenz
+              \end{enumerate}
+    \end{enumerate}
+    \vfill\null
+    \columnbreak
+
+\end{multicols*}