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								BA1 - Analyse I/BA1 - Analyse I.tex
									
									
									
									
									
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								BA1 - Analyse I/BA1 - Analyse I.tex
									
									
									
									
									
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\documentclass[fontsize=8pt, paper=a4, pagesize, DIV=calc]{scrartcl}
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\input{../Base.tex}
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\title{Formulaire d'Analyse I}
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\begin{document}
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\begin{tabu}to \textwidth{ |X|X| }
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\hline
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\textbf{Polynômes de Taylor} \newline
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$ \begin{aligned}
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                          \e^x & = \sum\limits_{k = 0}^\infty \frac{x^k}{k!}, &&x \in \symbb{R} \\
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       \sinh \left( x \right) & = \sum\limits_{k = 0}^\infty \frac{x^{2k+1}}{\left( 2k+1 \right)!}, &&x \in \symbb{R} \\
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       \cosh \left( x \right) & = \sum\limits_{k = 0}^\infty \frac{x^{2k}}{\left( 2k \right)!}, &&x \in \symbb{R} \\
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        \sin \left( x \right) & = \sum\limits_{k = 0}^\infty \left( -1 \right)^k \cdot \frac{x^{2k+1}}{\left( 2k+1 \right)!}, &&x \in \symbb{R} \\
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        \cos \left( x \right) & = \sum\limits_{k = 0}^\infty \left( -1 \right)^k \cdot \frac{x^{2k}}{\left( 2k \right)!}, &&x \in \symbb{R} \\
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       \ln \left( 1+x \right) & = \sum\limits_{k = 0}^\infty \left( -1 \right)^{k+1} \cdot \frac{x^k}{k}, &&x \in \left] -1, 1 \right[ \\
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                \frac{1}{1+x} & = \sum\limits_{k = 0}^\infty \left( -1 \right)^{k} \cdot x^k, &&x \in \left] -1, 1 \right[ \\
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     \arctan \left( x \right) & = \sum\limits_{k = 0}^\infty \left( -1 \right)^k \cdot \frac{x^{2k+1}}{2k+1}, &&x \in \left] -1, 1 \right[ \\
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\end{aligned} $
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&
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\textbf{Intégrales} \newline
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$ \begin{aligned}
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  &\int \frac{f' \left( x \right)}{f \left( x \right)} \cdot \dif x && = \ln \abs{f \left( x \right)} + C \\
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  &\int \frac{f' \left( x \right)}{1+f^2 \left( x \right)} \cdot \dif x && = \arctan \left[ f \left( x \right) \right] + C \\
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  &\int \left[ f \left( x \right) \right]^\alpha \cdot f' \left( x \right) \cdot \dif x && = \frac{\left[ f \left( x \right) \right]^{\alpha+1}}{\alpha + 1} + C, &\forall \alpha \neq -1 \\
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  &\int \e^{f \left( x \right)} \cdot f' \left( x \right) \cdot \dif x && = \e^{f \left( x \right)} + C \\
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  &\int \frac{f' \left( x \right)}{\sqrt{1-f^2 \left( x \right)}} \cdot \dif x && = \arcsin \left[ f \left( x \right) \right] + C \\
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\end{aligned} $
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\\
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\end{tabu}
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\nointerlineskip
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\begin{tabu}to \textwidth{ |X|X|X| }
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\hline
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\textbf{Racine carrée complexe} \newline
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$ \begin{aligned}
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  w = u + v \cdot \im, z = a + b \cdot \im, z^2 = w \\
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  \begin{cases}
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    a^2 - b^2 & = u \\
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    2 \cdot a \cdot b & = v \\
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    a^2 + b^2 & = \sqrt{u^2 + v^2} \\
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  \end{cases}
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\end{aligned} $
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&
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\textbf{Somme géométrique} \newline
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$ \begin{aligned}
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       \sum\limits_{k = 0}^n q^k & = \frac{1-q^{n+1}}{1-q} \\
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  \sum\limits_{k = 0}^\infty q^k & = \frac{1}{1-q} \\
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\end{aligned} $
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&
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\\\hline
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\textbf{Exponentielle} \newline
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$ \begin{aligned}
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   \cos \left( \theta \right) & = \frac{\e^{\im \cdot \theta} + \e^{-\im \cdot \theta}}{2} \\
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   \sin \left( \theta \right) & = \frac{\e^{\im \cdot \theta} - \e^{-\im \cdot \theta}}{2 \cdot \im} \\
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  \cosh \left( \theta \right) & = \frac{\e^{\theta} + \e^{-\theta}}{2} \\
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  \sinh \left( \theta \right) & = \frac{\e^{\theta} - \e^{-\theta}}{2} \\
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\end{aligned} $
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&
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\textbf{Exponentielle} \newline
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$ \begin{aligned}
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   \lim_{n \to \infty} \left( 1 + \frac{L}{n} \right)^n & = \e^L \\
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   \text{De manière\ générale~:} \\
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   \lim_{x \to \infty} f \left( x \right) & = +\infty \\
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   \lim_{x \to \infty} f \left( x \right) \cdot h \left( x \right) & = L \\
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   \lim_{x \to \infty} \left[ 1 + h \left( x \right) \right]^{f \left( x \right)} & = \e^L \\
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\end{aligned} $
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&
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\textbf{Trigonométrie} \newline
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$ \begin{aligned}
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   \cosh^2 \left( x \right) - \sinh^2 \left( x \right) = 1 \\
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   \cos^2 \left( x \right) + \sin^2 \left( x \right) = 1 \\
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   \sin \left( x+y \right) = \sin \left( x \right) \cdot \cos \left( y \right) + \cos \left( x \right) \cdot \sin \left( y \right) \\
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   \cos \left( x+y \right) = \cos \left( x \right) \cdot \cos \left( y \right) + \sin \left( x \right) \cdot \sin \left( y \right) \\
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   \sin \left( x \right) + \sin \left( y \right) = 2 \cdot \sin \left( \frac{x+y}{2} \right) \cdot \cos \left( \frac{x-y}{2} \right) \\
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   \sin \left( x \right) - \sin \left( y \right) = 2 \cdot \sin \left( \frac{x-y}{2} \right) \cdot \cos \left( \frac{x+y}{2} \right) \\
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   \cos \left( x \right) + \cos \left( y \right) = 2 \cdot \cos \left( \frac{x+y}{2} \right) \cdot \cos \left( \frac{x-y}{2} \right) \\
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   \cos \left( x \right) - \cos \left( y \right) = -2 \cdot \sin \left( \frac{x+y}{2} \right) \cdot \sin \left( \frac{x-y}{2} \right) \\
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\end{aligned} $
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\\\hline
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\end{tabu}
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\nointerlineskip
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\begin{tabu}to \textwidth{ |X| }
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\textbf{Angles particuliers} \newline
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$ \begin{aligned}
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   \cos \left( 0 \right) = 1 \qquad &\cos \left( \frac{\pi}{6} \right) = \frac{1}{2}         \qquad &\cos \left( \frac{\pi}{4} \right) = \frac{1}{2} \cdot \sqrt{2} \qquad &\cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \cdot \sqrt{3} \qquad &\cos \left( \frac{\pi}{2} \right) = 0 \\
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   \sin \left( 0 \right) = 0 \qquad &\sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \cdot \sqrt{3} \qquad &\sin \left( \frac{\pi}{4} \right) = \frac{1}{2} \cdot \sqrt{2} \qquad &\sin \left( \frac{\pi}{3} \right) = \frac{1}{2}         \qquad &\sin \left( \frac{\pi}{2} \right) = 1 \\
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\end{aligned} $
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\\\hline
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\textbf{Convergence} \newline
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$ \begin{aligned}
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  &\int_M^\infty x^a \cdot \e^{-bx} \cdot \dif x   &\text{ converge pour tout } a \in \symbb{R} \text{ et tout } b > 0 \\
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  &\int_a^\infty \frac{1}{x^p} \cdot \dif x &\text{ converge si et seulement si } p > 1 \quad \left( a > 0 \right) \\
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  &\int_0^b \frac{1}{x^p} \cdot \dif x      &\text{ converge si et seulement si } p < 1 \\
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\end{aligned} $
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\\\hline
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\end{tabu}
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\end{document}
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