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