Barre utilisateur

[ID: 716] [Date de publication: 18 janvier 2021 13:44] [Catégorie(s): Calcul de développements limités ] [ Nombre commentaires: 1] [nombre d'éditeurs: 1 ] [Editeur(s): Emmanuel Vieillard-Baron ] [nombre d'auteurs: 3 ] [Auteur(s): Alain Soyeur François Capaces Emmanuel Vieillard-Baron ]

Solution(s)

Solution(s)

Exercice 246
Par Alain Soyeur François Capaces Emmanuel Vieillard-Baron le 18 janvier 2021 13:44

1. Utilisant les formules usuelles : \[\begin{aligned} \ln\left({\scriptstyle x^2+1\over\scriptstyle x+1}\right)&=&\ln\left(1+x^2\right)-\ln\left(1+x\right)\\ &=&\boxed{-x+\dfrac{3}{2}x^2-\dfrac{x^3}{3}+ \underset{x \rightarrow 0}{o}\left(x^3\right)} \end{aligned}\]

2. Comme \(\tan x=x+\dfrac{x^3}{3}+\dfrac{2}{15}x^5+\underset{x \rightarrow 0}{o}\left(x^5\right)\) et que \(\operatorname{th} x=x-\dfrac{x^3}{3}+\dfrac{2}{15}x^5+\underset{x \rightarrow 0}{o}\left(x^5\right)\), on a : \[\begin{aligned} \dfrac{1}{\operatorname{th} x} - \dfrac{1}{\tan x}&=&\dfrac{1}{x-\dfrac{x^3}{3}+\dfrac{2}{15}x^5+\underset{x \rightarrow 0}{o}\left(x^5\right)} - \dfrac{1}{x+\dfrac{x^3}{3}+\dfrac{2}{15}x^5+\underset{x \rightarrow 0}{o}\left(x^5\right)}\\ &=& \dfrac{1}{x}\left( \dfrac{1}{1-\dfrac{x^2}{3}+\dfrac{2}{15}x^4+\underset{x \rightarrow 0}{o}\left(x^4\right)} - \dfrac{1}{1+\dfrac{x^2}{3}+\dfrac{2}{15}x^4+\underset{x \rightarrow 0}{o}\left(x^4\right)} \right)\\ &=& \dfrac{1}{x}\left(1+\dfrac{x^2}{3} -\dfrac{x^4}{45}- 1+\dfrac{x^2}{3}+ \dfrac{x^4}{45}+\underset{x \rightarrow 0}{o}\left(x^4\right)\right) \\ &=& \boxed{\dfrac{2}{3}x+\underset{x \rightarrow 0}{o}\left(x^3\right)} \end{aligned}\]

3. \[\begin{aligned} \sqrt{1+\cos x}&=&\sqrt{1+1-\dfrac{x^2}{2}+\dfrac{x^4}{24}+\underset{x \rightarrow 0}{o}\left(x^5\right)}\\ &=&\sqrt{2}\sqrt{1 -\dfrac{x^2}{4}+\dfrac{x^4}{48}+\underset{x \rightarrow 0}{o}\left(x^5\right) }\\ &=&\boxed{\sqrt{2}\left(1-\dfrac{x^2}{8}+\dfrac{x^4}{384}\right)+\underset{x \rightarrow 0}{o}\left(x^5\right)} \end{aligned}\]

4. \[\begin{aligned} \left(1+x\right)^{{\scriptstyle 1\over\scriptstyle x}}&=&e^{{\scriptstyle\ln\left(1+x\right)\over\scriptstyle x}}\\ &=& e^{\dfrac{1}{x}\left(x-\dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} +\underset{x \rightarrow 0}{o}\left(x^4\right)\right)}\\ &=& e^{1-\dfrac{x}{2} + \dfrac{x^2}{3}-\dfrac{x^3}{4}+ \underset{x \rightarrow 0}{o}\left(x^3\right) }\\ &=&e e^{ -\dfrac{x}{2} + \dfrac{x^2}{3}-\dfrac{x^3}{4}+ \underset{x \rightarrow 0}{o}\left(x^3\right) }\\ &=&\boxed{e\left(1-\dfrac{x}{2} + \dfrac{11}{24}x^2 -\dfrac{7}{16}x^3 \right)+ \underset{x \rightarrow 0}{o}\left(x^3\right)} \end{aligned}\]

5. \[\begin{aligned} \ln\left({\scriptstyle\mathop{\mathrm{sh}}x\over\scriptstyle x}\right)&=&\ln\left(\dfrac{1}{x}\left(x+\dfrac{x^3}{6} +\dfrac{x^5}{120}+\underset{x \rightarrow 0}{o}\left(x^5\right)\right)\right)\\ &=&\ln\left(1+ \dfrac{x^2}{6} +\dfrac{x^4}{120}+\underset{x \rightarrow 0}{o}\left(x^4\right) \right)\\ &=& \boxed{\dfrac{x^2}{6}-\dfrac{x^4}{180}+\underset{x \rightarrow 0}{o}\left(x^4\right)} \end{aligned}\]

6. \[\begin{aligned} \ln\left(1+\sqrt{1+x}\right)&=& \ln\left(2+\dfrac{x}{2}-\dfrac{x^2}{8} + \dfrac{x^3}{16} + \underset{x \rightarrow 0}{o}\left(x^3\right) \right)\\ &=& \ln2\left(1+\dfrac{x}{4}-\dfrac{x^2}{16} + \dfrac{x^3}{32} + \underset{x \rightarrow 0}{o}\left(x^3\right) \right)\\ &=& \ln 2 + \ln \left(1+\dfrac{x}{4}-\dfrac{x^2}{16} + \dfrac{x^3}{32} + \underset{x \rightarrow 0}{o}\left(x^3\right)\right)\\ &=& \boxed{\ln 2 + \dfrac{x}{4}-\dfrac{3}{32}x^2+\dfrac{5}{96}x^3+ \underset{x \rightarrow 0}{o}\left(x^3\right)} \end{aligned}\]