{"id":34245,"date":"2019-08-19T11:37:04","date_gmt":"2019-08-19T10:37:04","guid":{"rendered":"https:\/\/groupe-reussite.fr\/?page_id=34245"},"modified":"2022-07-20T11:27:03","modified_gmt":"2022-07-20T10:27:03","slug":"cours-en-ligne-corriges-taylor-developpements-limites-ecs1","status":"publish","type":"page","link":"https:\/\/groupe-reussite.fr\/ressources\/cours-en-ligne-corriges-taylor-developpements-limites-ecs1\/","title":{"rendered":"Formules de Taylor et d\u00e9veloppements limit\u00e9s en ECG1 – Corrig\u00e9s"},"content":{"rendered":"
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Cours en ligne Maths en ECG1<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en ECG1<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFormules de Taylor et D\u00e9veloppements Limit\u00e9s<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Cours en ligne Maths en ECG1<\/a><\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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Chapitres Maths en ECG1<\/p>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tProbabilit\u00e9s (univers infini)<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tFormules de Taylor et D\u00e9veloppements Limit\u00e9s<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Exercices corrig\u00e9s : Formules de Taylor, d\u00e9veloppements limit\u00e9s ECG1<\/h1>\n<\/div>\n

R\u00e9sum\u00e9 de cours<\/a> Exercices<\/a> Corrig\u00e9s<\/a><\/p>\n

Cours en ligne de Maths en ECG1<\/a><\/p>\n

Corrig\u00e9s – Formules de Taylor et d\u00e9veloppements limit\u00e9s<\/h2>\n

Exercice 1 :\u00a0<\/b><\/h3>\n

1)<\/strong> On pose \"u = \dfrac{x}{2}.\" On a :<\/p>\n

\"\sqrt{1 + u} = \left( 1 + u \right)^{1 / 2}\"
\n\"\underset{u \to 0}{=} 1 + \dfrac12 u - \dfrac18 u^2 + \dfrac{1}{16} u^3 + o \left( u^3 \right)\"
\n\"\underset{x \to 0}{=} 1 + \dfrac14 x - \dfrac{1}{32} x^2 + \dfrac{1}{128} x^3 + o \left( x^3 \right)\".<\/p>\n

2)<\/strong> On pose \"x = \dfrac{\pi}{4} + h.\" Faisons le d\u00e9veloppement limit\u00e9 de \"\sin \left( \dfrac{\pi}{4} + h \right)\" \u00e0 l’ordre \"5\" lorsque \"h\" tend vers \"0.\" On a<\/p>\n

\"\sin \left( \dfrac{\pi}{4} + h \right)\" \"= \dfrac{\sqrt{2}}{2} \left( \cos \left( h \right) + \sin \left( h \right) \right)\"
\n\"\underset{h \to 0}{=} \dfrac{\sqrt{2}}{2} ( 1 - \dfrac{h^2}{2!} + \dfrac{h^4}{4!} + o \left( h^5 \right) h\" \"- \dfrac{h^3}{3!} + \dfrac{h^5}{5!} + o \left( h^5 \right))\"
\n\"\underset{h \to 0}{=} \dfrac{\sqrt{2}}{2} ( 1 + h - \dfrac{h^2}{2} - \dfrac{h^3}{6}\" \"+ \dfrac{h^4}{24} + \dfrac{h^5}{120} + o \left( h^5 \right) )\".<\/div>\n
Comme \"h = x - \dfrac{\pi}{4},\" on a finalement,<\/div>\n
\"\sin \left( x \right) \underset{x \to \dfrac{\pi}{4}}{=}\" \"\dfrac{\sqrt{2}}{2} ( 1 + \left( x - \dfrac{x}{4} \right) - \dfrac{\left( x - \dfrac{x}{4} \right)^2}{2}\" \"- \dfrac{\left( x - \dfrac{x}{4} \right)^3}{6} + \dfrac{\left( x - \dfrac{x}{4} \right)^4}{24}\" \"+ \dfrac{\left( x - \dfrac{x}{4} \right)^5}{120}\" \"+ o \left( \left( x - \dfrac{x}{4} \right)^5 \right) ).\"
\nOn ne d\u00e9veloppe pas, bien s\u00fbr !<\/div>\n
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3)<\/strong> On a \"e^{-x} \underset{x \to 0}{=}\"\u00a0 \"1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{6} + o \left( x^3 \right)\" et \"\cos \left( 2 x \right) \underset{x \to 0}{=}\" \"1 - \dfrac{\left( 2 x \right)^2}{2!} + o \left( x^3 \right).\" Ce qui donne :<\/p>\n

\"e^{-x} \cos\left( 2 x \right) \underset{x \to 0}{=}\" \"\left( 1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{6} + o \left( x^3 \right) \right)\" \"\times \left( 1 - \dfrac{\left( 2 x \right)^2}{2!} + o \left( x^3 \right) \right).\"
\nLorsque que l’on d\u00e9veloppe, les termes ayant \"x^k\" avec \"k \ge 4\" \u00ab\u00a0entrant\u00a0\u00bb dans \"o \left( x ^3 \right).\" Apr\u00e8s calculs, il vient que :<\/div>\n
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\"e^{-x} \cos\left( 2 x \right) \underset{x \to 0}{=}\" \"1 - 3 x + \dfrac52 x^2 - \dfrac76 x^3 + o \left( x^3 \right).\"<\/div>\n
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4)<\/strong> On commence par \u00e9crire :<\/span><\/div>\n
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\"\dfrac{\cos \left( x \right)}{2 - x}\" \"= \dfrac12 \dfrac{\cos \left( x \right)}{1 - \dfrac{x}{2}}\"<\/p>\n

\"= \dfrac12 \cos \left( x \right) \times \left( 1 - \dfrac{x}{2} \right)^{-1}\"
\n\"\underset{x \to 0}{=} \dfrac12 \left( 1 - \dfrac{x^2}{2} + o \left( x^3 \right) \right)\" \"\times \left( 1 + \dfrac12 x + \dfrac14 x^2 + \dfrac18 x^3 + o \left( x^3 \right) \right)\"
\n\"\underset{x \to 0}{=} \dfrac12 ( 1 + \dfrac12 x + \dfrac14 x^2 + \dfrac18 x^3\" \"- \dfrac12 x^2 - \dfrac14 x^3 + o \left( x^3 \right) )\"
\n\"\underset{x \to 0}{=} \dfrac12 \left( 1 + \dfrac12 x - \dfrac14 x^2 - \dfrac18 x^3 + o \left( x^3 \right) \right)\".<\/div>\n<\/div>\n<\/div>\n

 <\/p>\n

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COURS DE MATHS<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Les meilleurs professeurs particuliers<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Pour progresser et r\u00e9ussir<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCours particuliers maths<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Avis Google France <\/b>\u2605\u2605\u2605\u2605\u2605<\/b><\/span> 4,9 sur 5<\/b><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\n

 <\/p>\n

Exercice 2 :\u00a0<\/b><\/h3>\n

1) On a \"\ln \left( 1 + x \right) \underset{x \to 0}{=}\" \"x - \dfrac{x^2}{2} + o \left( x^2 \right).\" Ainsi \"\ln \left( 1 + x \right) - x \underset{x \to 0}{=}\" \"- \dfrac12 x^2 + o \left( x^2 \right)\" soit \"\ln \left( 1 + x \right) - x \underset{x \to 0}{\sim} - \dfrac12 x^2.\"<\/p>\n

Ainsi \"\dfrac{\ln \left( 1 +x \right) - x}{x^2} \underset{x \to 0}{\sim} \dfrac{- \dfrac12 x^2}{x^2} \underset{x \to 0}{\sim} - \dfrac12 .\" En particulier,<\/p>\n

\n

  <\/span>   <\/span>\"\[\lim_{x \to 0} \dfrac{\ln \left( 1 + x \right) - x}{x^2} = - \dfrac12.\]\"<\/p>\n<\/div>\n

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2) Trouvons un \u00e9quivalent du num\u00e9rateur et du d\u00e9nominateur.<\/span><\/p>\n

<\/div>\n
Num\u00e9rateur :<\/i><\/div>\n
<\/div>\n
On a \"e^x \underset{x \to 0}{=} 1 + x + o \left( x \right).\" Donc<\/div>\n
<\/div>\n
\"\left( x - 1 \right) e^x + 1 \underset{x \to 0}{=}\" \"\left( x - 1 \right) \left( 1 + x + \dfrac12 x^2 + o \left( x^2 \right) \right) + 1\" \"\underset{x \to 0}{=} x^2 + o \left( x^2 \right).\"
\nCe qui donne<\/div>\n
<\/div>\n
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  <\/span>   <\/span>\"\[\left( x - 1 \right) e^x + 1 \underset{x \to 0}{\sim} x^2 .\]\"<\/p>\n<\/div>\n

<\/div>\n
D\u00e9nominateur :<\/i><\/div>\n
<\/div>\n
On a \"e^x \underset{x \to 0}{=} 1 + x + o \left( x \right)\" soit \"e^x - 1 \underset{x \to 0}{\sim} x .\" Ainsi<\/div>\n
<\/div>\n
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  <\/span>   <\/span>\"\[x \left( e^x - 1 \right) \underset{x \to 0}{\sim} x^2.\]\"<\/p>\n

Finalement<\/p><\/div>\n

<\/div>\n
\"\dfrac{\left( x - 1 \right) e^x + 1}{x \left( e^x - 1 \right)}\" \"\underset{x \to 0}{\sim} \dfrac{x^2}{x^2}\"<\/div>\n
\"\underset{x \to 0}{\sim} 1\".<\/div>\n
Et \"\lim_{x \to 0} \dfrac{\left( x - 1 \right) e^x + 1}{x \left( e^x - 1 \right)} = 1.\"<\/div>\n
<\/div>\n
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3) Trouvons un \u00e9quivalent du num\u00e9rateur et du d\u00e9nominateur.<\/span><\/p>\n

<\/div>\n
Num\u00e9rateur :<\/i><\/div>\n
<\/div>\n
On a<\/div>\n
\"e^x - 1 - \sin \left( x \right) \underset{x \to 0}{=}\" \"\left( 1 + x + \dfrac12 x^2 + o \left( x^2 \right) \right) - 1 - \left( x + o \left( x^2 \right) \right)\"
\n\"\underset{x \to 0}{=} \dfrac12 x^2 + o \left( x^2 \right)\".<\/div>\n
Ce qui donne<\/div>\n
<\/div>\n
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  <\/span>   <\/span>\"\[e^x - 1 - \sin \left( x \right) \underset{x \to 0}{\sim} \dfrac12 x^2.\]\"<\/p>\n<\/div>\n

<\/div>\n
D\u00e9nominateur :<\/i><\/div>\n
<\/div>\n
On a \"\cos \left( x \right) \underset{x \to 0}{=} 1 - \dfrac12 x^2 + o \left( x^2 \right),\" soit<\/div>\n
<\/div>\n
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  <\/span>   <\/span>\"\[\cos \left( x \right) - 1 \underset{x \to 0}{\sim} - \dfrac12 x^2 .\]\"<\/p>\n

Ce qui donne finalement :<\/p><\/div>\n

\"\dfrac{e^x - 1 - \sin \left( x \right)}{\cos \left( x \right) - 1}\" \"\underset{x \to 0}{\sim} \dfrac{\dfrac12 x^2}{- \dfrac12 x^2}\"
\n\"\underset{x \to 0}{\sim} - 1\".<\/div>\n
Ainsi<\/div>\n
<\/div>\n
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  <\/span>   <\/span>\"\[\lim_{x \to 0 } \dfrac{e^x - 1 - \sin \left( x \right)}{\cos \left( x \right) - 1} = - 1.\]\"<\/p>\n<\/div>\n

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Exercice 3 :\u00a0<\/b>
\nFaisons un d\u00e9veloppement limit\u00e9 \u00e0 l’ordre \"4\" du num\u00e9rateur. On a\"\sqrt{1 + 2x} \underset{x \to 0}{=}\" \"1 + x - \dfrac12 x^2 + \dfrac12 x^3 - \dfrac58 x^4 + o \left( x^4 \right),\"<\/p>\n
\"\cos \left( x \right) \underset{x \to 0}{=}\" \"1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} + o \left( x^4 \right)\" \"= 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} + o \left( x^4 \right).\"
\nAinsi<\/div>\n
\"\sqrt{1 + 2x } - x - \cos \left( x \right) \underset{x \to 0}{=}\" \"\dfrac12 x^3 - \dfrac{7}{12} x^4 + o \left( x^4 \right).\"
\nCe qui donne finalement :<\/div>\n
\"\dfrac{\sqrt{1 + 2x } - x - \cos \left( x \right)}{x^3} \underset{x \to 0}{=}\" \"\dfrac12 - \dfrac{7}{12} x + o \left( x \right).\"
\nOn a donc montr\u00e9 que \"f\" admet un d\u00e9veloppement limit\u00e9 \u00e0 l’ordre \"1\" en \"0,\" il s’ensuit que \"f\" est d\u00e9rivable en \"0\" et \"f ' \left( 0 \right) = - \dfrac{7}{12}.\" En particulier, \"f\" est continue en \"0.\"<\/div>\n
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COURS PARTICULIERS EN LIGNE<\/h3>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Nous avons s\u00e9lectionn\u00e9 pour vous les meilleurs professeurs particuliers.<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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POUR ACC\u00c9L\u00c9RER MA PROGRESSION<\/h4>\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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\n\t\t\t\n\t\t\t\t\t\t\n\t\t\t\t\t\tCours particuliers en ligne<\/span>\n\t\t<\/span>\n\t\t\t\t\t<\/a>\n\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t
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Avis Google France <\/b>\u2605\u2605\u2605\u2605\u2605<\/b><\/span> 4,9 sur 5<\/b><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t

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\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t<\/div>\n
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Exercice 4 :\u00a0<\/strong><\/h3>\n

On utilise la formule de Taylor-Young \u00e0 \"f\" \u00e0 l’ordre \"2\" en \"0\" :<\/p>\n

\"f \left( x \right) \underset{x \to 0}{=}\" \"f \left( 0 \right) + f ' \left( 0 \right) x + \dfrac{f'' \left(0 \right)}{2} x^2 + o \left( x^2 \right).\"
\nAinsi,<\/div>\n
\"\dfrac{f \left( x \right) - f \left( 0 \right)}{x} \underset{x \to 0}{=}\" \"f ' \left( 0 \right) + \dfrac{f'' \left( 0 \right)}{2} x + o \left( x \right).\"
\nDe plus,<\/div>\n
\"f ' \left( x \right) \underset{x \to 0}{=}\" \"f ' \left( 0 \right) + f'' \left( 0 \right) x + o \left( x \right).\"
\nSoit finalement<\/div>\n
\"f ' \left( x \right) - \dfrac{f \left( x \right) - f \left( 0 \right)}{x} \underset{x \to 0}{=}\" \"\dfrac12 f '' \left( 0 \right) x + o \left( x \right),\"
\nce qui donne<\/div>\n
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\"\lim_{x \to 0} \dfrac{f' \left( x \right) - \dfrac{f \left( x \right) - f \left( 0 \right)}{x}}{x}\" \"= \dfrac12 f '' \left( 0 \right).\"<\/p>\n

N’attendez pas le dernier moment pour r\u00e9viser les maths en ECG1. Faites la diff\u00e9rence avec vos camarades en prenant de l’avance sur les chapitres de maths \u00e0 venir :<\/p>\n<\/div>\n

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