\documentclass[11pt,fleqn]{scrartcl} \usepackage{ucs} \usepackage[utf8x]{inputenc} \usepackage{ngerman} \usepackage{graphicx} \usepackage{amsmath,amssymb,amstext,bbm} \usepackage[automark]{scrpage2} \title{Formelheft} \author{Helmuth Peer} \date{\today{}, Weiz} \begin{document} \maketitle \thispagestyle{empty}% weil \maketitle ggf. ein \thispagestyle{plain} enthält \tableofcontents %\pagestyle{empty} %\ifoot[]{Peer} %\cfoot{} %\ofoot{} \section{Potenzen} $a, b \in \mathbb{R}, r, s \in \mathbb{R}, k \in \mathbb{Z}, m, n \in \mathbb{N}^{\ast}$ $a^0 = 1$ $a^{- n} = \frac{1}{a^n} = \left( \frac{1}{a} \right)^n$ $a^{\frac{1}{n}} = \sqrt[n]{a}$ $a^r \cdot a^s = a^{r + s}$ $\sqrt[n]{a^k} = \sqrt[n \cdot m]{a^{k \cdot m}}$ $( \sqrt[n]{a})^k = \sqrt[n]{a^k}$ $(a \pm b)^2 = a^2 \pm 2 ab + b^2$ \section{Logarithmen} $a, b \in \mathbbm{R}^+ \backslash \left\{ 1 \right\}, u, v \in \mathbbm{R}^+, r \in \mathbbm{R}, n \in \mathbbm{N}^{\ast} \bot \in \backslash$ \[e = \lim_{n \rightarrow \infty}^{} \left( 1 + \frac{1}{n} \right)^n\] \[ \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n = 2, 71828 \ldots \] \section{Quadratische Gleichungen} \[x^2 + px + q = 0\] \[x = - \frac{p}{2} \pm \sqrt{\left( \frac{p}{2} \right)^2 - q}\] \begin{tabular}{|l|l|l|} \hline $x^2 + px + q = 0$ & & $ax^2 + bx + c = 0$\\ \hline $x = - \frac{p}{2} \pm \sqrt{\left( \frac{p}{2} \right)^2 - q}$ & & $x = \frac{- b \pm \sqrt{b^2 - 4 ac}}{2 a}$\\ \hline \end{tabular} \section{Komplexe Zahlen} $z = a + bi \in \mathbbm{C} \Leftrightarrow a, b \in \mathbbm{R} \ \mbox{und} \ i^2 = - 1 ; \sqrt{- a} = i \sqrt{a} \ \mbox{mit} \ a > 0$ \\ $(a + bi) (a - bi) = a^2 + b^2$ $\left| z \right| = r = \sqrt{a^2 + b^2}$ $\arg z = \varphi \in [0^{\circ} ; 360^{\circ} [$ %\includegraphics{oelta002.png} \section{Schaltalgebra} $a \vee b = b \vee a$ $a \wedge b = b \wedge a$ \section{Vektoren} \subsection{Vektorielles Produkt} $\vec{a} = \left(\begin{array}{c} a_1\\ a_2\\ a_3 \end{array}\right) \times \left(\begin{array}{c} b_1\\ b_2\\ b_3 \end{array}\right) = \left(\begin{array}{c} \left|\begin{array}{c} a_2 b_2\\ a_3 b_3 \end{array}\right|\\ - \left|\begin{array}{c} a_1 b_1\\ a_3 b_3 \end{array}\right|\\ \left|\begin{array}{c} a_1 b_1\\ a_2 b_2 \end{array}\right| \end{array}\right) = \left(\begin{array}{c} a_2 b_3 - a_3 b_2\\ a_3 b_1 - a_1 b_3\\ a_1 b_2 - a_2 b_1 \end{array}\right)$ \section{Analytische Geometrie} $\overrightarrow{\text{AB}} = B - A$ \subsection{Fl\"acheninhalt Parallelogramm} $A_p = \sqrt{\vec{a}^2 \cdot \vec{b}^2 - ( \vec{a} \cdot \vec{b})^2}$ \subsection{Parameterdarstellung einer Geraden} $\vec{x} = \left(\begin{array}{c} 1\\ 2\\ 3 \end{array}\right) + t \cdot \left(\begin{array}{c} 4\\ 5\\ 6 \end{array}\right)$ \section{Differential- und Integralrechnung} \subsection{Ableitungs- und Stammfunktionen} \begin{tabular}{|l|l|l|} \hline Funktion & Ableitungsfunktion & Stammfunktionen\\ \hline $y = f (x) = k$ & $y' = f' (x) = 0$ & $F (x) = \int \mbox{kdx} = \mbox{kx} + C$\\ \hline $y = f (x) = x^q$ & $y' = f' (x) = q \cdot x^{q - 1}$ & $q \neq - 1 :$ \\ \hline & & $F (x) = \int x^q \mbox{dx} = \frac{x^{q + 1}}{q + 1} + C$\\ \hline \end{tabular} \subsection{Rauminhalte} \subsubsection{Drehk\"orper} Drehung um die x-Achse: \(V = \pi \int^b_a y^2 \mbox{dx}\) \subsection{Numerische Integration} \subsubsection{Rechtecksformel} $\int^b_a f (x) \mbox{dx} \approx \frac{b - a}{n} \cdot [f (x_0) + f (x_1) + f (x_2) + \ldots . + f (x_{n - 1})] = \Delta x \cdot \sum^{n - 1}_{i = 0} f (x_i)$ \[ \Delta x \cdot \sum_{i = 0}^{i - 1} f (x_i) \] \subsection{Binomialverteilung} $P (X = k) = b_{n, p} (k) = \left(\begin{array}{c} n\\ k \end{array}\right) p^k (1 - p)^{n - k}$ \subsection{Normalverteilung} $\varphi (x) = \frac{1}{\sqrt{2 \pi}} e^{- \frac{x^2}{2}}$ \[ \Phi (z) = \int^z_{- \infty} \varphi (x) \text{dx} = \frac{1}{\sqrt{2 \pi}} \int^z_{- \infty} e^{- \frac{x^2}{2}} \mbox{dx} \] \end{document}
http://de.wikibooks.org/wiki/LaTeX-Kompendium:_F%C3%BCr_Mathematiker http://mirror.ctan.org/info/symbols/comprehensive/symbols-a4.pdf
\documentclass[11pt,fleqn]{scrartcl} \usepackage{ucs} \usepackage[utf8x]{inputenc} \usepackage{ngerman} \usepackage{graphicx} \usepackage{amsmath,amssymb,amstext,bbm} \usepackage[automark]{scrpage2} \usepackage{cancel} \title{Formelheft} \author{Helmuth Peer} \date{\today{}, Weiz} \begin{document} \maketitle \thispagestyle{empty}% weil \maketitle ggf. ein \thispagestyle{plain} enthält \tableofcontents %\pagestyle{empty} %\ifoot[]{Peer} %\cfoot{} %\ofoot{} \section{Potenzen} $a, b \in \mathbbm{N}^{\ast}$ $x^0=1$ $a^{-n}=\frac{1}{a^n}=\left(\frac{1}{a}\right)^n$ $a^{\frac{1}{n}}=\sqrt[n]{a}$ $a^r \cdot a^s=a^{r+s}$ \section{Logarithmen} $a, b \in \mathbb{R}^+ \backslash \left\{1 \right\}$ \[e=\lim_{n \rightarrow \infty} \left(1+ \frac{1}{n} \right)^n\] \section{Approximation} X ist binomialverteilt mit den Parametern $\mu=np \mbox{ und } \sigma=\sqrt{np(1-p)}$ Für alle Zahlen \(a_1 \dots a_n\) gilt \dots %Formel in der Textzeile \(\frac{13}{39} = \cancel{{\frac{13}{13}}} \cdot \frac{1}{3}\) \(\frac{13}{39} = \frac{\cancel{13} \cdot 1}{\cancel{13} \cdot 3}\) \(\frac{13}{39} = \cancelto{1}{\frac{13}{13}} \frac{1}{3}\) \section{Weitere Verteilungen} \subsection{Hypergeometrische Verteilung} \(n,N,M \in \mathbb{N}^{\ast}, k \in \mathbb{N}, n \le N, k \le n, k \le M, n-k \le N, M \le N\) \[P(X=k)= \frac{\left(\begin{array}{c} M\\ k \end{array} \right) \left(\begin{array}{c} N-M\\ n-k \end{array} \right)} {\left(\begin{array}{c} N\\ n \end{array} \right)} \] \[\mu=E(X)=n \cdot \frac{M}{N} \hspace*{2.5cm} \sigma^2=V(X)=n \cdot \frac{M}{N} \cdot \left(1-\frac{M}{N}\right) \cdot \frac{N-n}{N-1} \] \subsection{Poisson-Verteilung} \(n \in \mathbb{N}^{\ast},0 \le p \le 1, \mu=n \cdot p \) \(P(X=x)=f(x)=\frac{\mu^x}{x!} \cdot e^{-\mu} \hspace{1.5cm} \mu=E(X)=n \, p \hspace{1.5cm} \sigma^2=V(X)=n\,p\) \section{Chi-Quadrat-Abweichungsmaß} \(b_i \dots \mbox{beobachtete Häufigkeit in der Klasse i}\, (i=1,2, \dots , n)\) \(e_i \dots \mbox{erwartete Häufigkeit in der Klasse i}\) \[\chi^2=\frac{(b_1-e_1)^2}{e_1}+\frac{(b_2-e_2)^2}2{e_2}+ \dots + \frac{(b_n-e_n)^2}{e_n}=\sum_{i=1}^{n} \frac{(b_i-e_i)^2}{e_i}\] \end{document}