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added some deriviates

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Martin Thoma 2012-09-16 12:38:49 +02:00
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cheat-sheets/analysis/Analysis_Wichtige_Formeln.tex
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@ -1,6 +1,7 @@
\documentclass[a4paper,10pt]{article}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amssymb, amsmath}
\DeclareMathOperator{\arcsinh}{arcsinh}
\DeclareMathOperator{\arccosh}{arccosh}
\DeclareMathOperator{\arctanh}{arctanh}
\usepackage[utf8]{inputenc} % this is needed for umlauts
\usepackage[ngerman]{babel} % this is needed for umlauts
@ -30,12 +31,12 @@
\begin{minipage}[b]{0.5\linewidth}\centering
\begin{align*}
\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)!}
\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)!}
\end{align*}
\end{minipage}
@ -55,30 +56,47 @@ e^x &= \sum_{n = 0}^{\infty} \frac {x^n}{n!} \\
\end{minipage}
\end{table}
\section{Zusammenhänge}
\begin{align*}
(\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} \\
(\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} \\
(x + y)^n &= \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\end{align*}
\section{Ableitungen}
\begin{table}[ht]
\begin{minipage}[b]{0.5\linewidth}\centering
\begin{align*}
(\arctan x)' &= \frac {1}{1 + x^2} \\
(\sin x)' &= \cos x \\
(\cos x)' &= -\sin x \\
(\arctanh x)' &= \frac {1}{1 - x^2}
(\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
\begin{align*}
(\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}
\end{align*}
\end{minipage}
\end{table}
\section{Potenzreihen}
Zuerst den Potenzradius r berechnen:
\(
r = \frac {1}{\lim \text{sup} \sqrt[n]{|a_n|}}
r = \frac {1}{\lim \text{sup} \sqrt[n]{|a_n|}}
\)
\end{document}