2
0
Fork 0
mirror of https://github.com/MartinThoma/LaTeX-examples.git synced 2025-04-26 06:48:04 +02:00
LaTeX-examples/cheat-sheets/analysis/Analysis_Wichtige_Formeln.tex

108 lines
3 KiB
TeX
Raw Normal View History

2012-08-10 10:56:38 +02:00
\documentclass[a4paper,10pt]{article}
2012-09-16 12:38:49 +02:00
\usepackage{amssymb, amsmath}
\DeclareMathOperator{\arcsinh}{arcsinh}
\DeclareMathOperator{\arccosh}{arccosh}
2012-09-16 12:10:10 +02:00
\DeclareMathOperator{\arctanh}{arctanh}
2012-08-10 10:56:38 +02:00
\usepackage[utf8]{inputenc} % this is needed for umlauts
\usepackage[ngerman]{babel} % this is needed for umlauts
\usepackage[T1]{fontenc} % this is needed for correct output of umlauts in pdf
%layout
\usepackage[margin=2.5cm]{geometry}
\usepackage{parskip}
\pdfinfo{
/Author (Peter Merkert, Martin Thoma)
/Title (Wichtige Formeln der Analysis I)
/CreationDate (D:20120221095400)
/Subject (Analysis I)
/Keywords (Analysis I; Formeln)
}
\everymath={\displaystyle}
\begin{document}
\title{Analysis Formelsammlung}
\author{Peter Merkert, Martin Thoma}
\date{21. Februar 2012}
\section{Grenzwerte}
\begin{table}[ht]
\begin{minipage}[b]{0.5\linewidth}\centering
\begin{align*}
2012-09-16 12:38:49 +02:00
\lim_{x \to 0} \frac {\sin x}{x} &= 1 \\
\lim_{x \to 0} \frac {e^x - 1}{x} &= 1 \\
\lim_{h \to 0} \frac {e^{{x_0} + h} - e^{x_0}}{h} &= e^{x_0} \\
\sum_{n = 0}^{\infty} (-1)^n \frac {(-1)^{n + 1}}{n} &= \log 2 \\
\cos x &= \sum_{n = 0}^{\infty} (-1)^n \frac {x^{2n}}{(2n)!} \\
\sin x &= \sum_{n = 0}^{\infty} (-1)^n \frac {x^{2n + 1}}{(2n + 1)!}
2012-08-10 10:56:38 +02:00
\end{align*}
\end{minipage}
\hspace{0.5cm}
\begin{minipage}[b]{0.5\linewidth}
\centering
\begin{align*}
\cosh x = \frac {1}{2} (e^x + e^{-x}) &= \scriptstyle \sum_{n = 0}^{\infty} \frac {x^{2n}}{(2n)!} \\
\sinh x = \frac {1}{2} (e^x - e^{-x}) &= \sum_{n = 0}^{\infty} \frac {x^{2n + 1}}{(2n + 1)!} \\
e^x &= \sum_{n = 0}^{\infty} \frac {x^n}{n!} \\
\sum_{n = 0}^{\infty} (-1)^n \frac {x^{n + 1}}{n + 1} &= \log (1+x) (x \in (-1,1)) \\
\sum_{n = 0}^{\infty} x^n &= \frac {1}{1 - x} (x \in (-1,1)) \\
0,\bar{3} &= \sum_{n = 1}^{\infty} \frac {3}{(10)^n}
\end{align*}
\end{minipage}
\end{table}
\section{Zusammenhänge}
\begin{align*}
2012-09-16 12:38:49 +02:00
(\cos x)^2 + (\sin x)^2 &= 1 \\
(\cosh x)^2 - (\sinh x)^2 &= 1 \\
\tan x &= \frac {\sin x}{\cos x} \\
\tanh x &= \frac {\sinh x}{\cosh x} \\
2012-08-10 10:56:38 +02:00
(x + y)^n &= \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\end{align*}
\section{Ableitungen}
\begin{align*}
(\log x)' &= \frac{1}{x}
\end{align*}
2012-09-16 12:38:49 +02:00
\begin{table}[ht]
\begin{minipage}[b]{0.5\linewidth}\centering
\begin{align*}
(\sin x)' &= \cos x \\
(\cos x)' &= -\sin x \\
(\tan x)' &= \frac{1}{\cos^2 x} \\
(\sinh x)' &= \cosh x \\
(\cosh x)' &= \sinh x \\
\end{align*}
\end{minipage}
\hspace{0.5cm}
\begin{minipage}[b]{0.5\linewidth}
\centering
2012-08-10 10:56:38 +02:00
\begin{align*}
2012-09-16 12:38:49 +02:00
(\arcsin x)' &= \frac {1}{\sqrt{1-x^2}} \\
(\arccos x)' &= - \frac {1}{\sqrt{1-x^2}} \\
(\arctan x)' &= \frac {1}{1 + x^2} \\
% (\arcsinh x)' &= \frac {1}{\sqrt{1+x^2}} \\
% (\arccosh x)' &= \frac {1}{\sqrt{(1-x^2) \cdot (1+x^2)}} \\
% (\arctanh x)' &= \frac {1}{1 - x^2}
2012-08-10 10:56:38 +02:00
\end{align*}
2012-09-16 12:38:49 +02:00
\end{minipage}
\end{table}
2012-08-10 10:56:38 +02:00
\section{Potenzreihen}
Zuerst den Potenzradius r berechnen:
\(
2012-09-16 12:38:49 +02:00
r = \frac {1}{\lim \text{sup} \sqrt[n]{|a_n|}}
2012-08-10 10:56:38 +02:00
\)
\end{document}