{"id":238548,"date":"2022-03-30T17:07:37","date_gmt":"2022-03-30T15:07:37","guid":{"rendered":"https:\/\/sherpas.com\/blog\/?p=238548"},"modified":"2024-08-28T22:48:55","modified_gmt":"2024-08-28T20:48:55","slug":"analyse-asymptotique-exercice-corrige","status":"publish","type":"post","link":"https:\/\/sherpas.com\/blog\/analyse-asymptotique-exercice-corrige\/","title":{"rendered":"Analyse asymptotique : exercice corrig\u00e9"},"content":{"rendered":"\n

Besoin d’aide sur un exercice portant sur l’analyse asymptotique<\/strong> ? Pas de panique ! Gr\u00e2ce \u00e0 ce cours d\u00e9di\u00e9 \u00e0 la notion : analyse asymptotique : exercice corrig\u00e9<\/strong>, entra\u00eene-toi avec des m\u00e9thodologies bien rod\u00e9es et r\u00e9ussis \u00e0 coup s\u00fbr tes prochaines interrogations \u00e9crites et orales sur le sujet !<\/p>\n

Optimise ton temps de r\u00e9vision gr\u00e2ce \u00e0 un prof particulier de maths en ligne<\/strong><\/a> sp\u00e9cialis\u00e9 en exercices d’analyse asymptotique. \ud83d\udd52<\/p>\n\n\n\n

Exercice : Analyse asymptotique<\/h2>\n\n\n\n

\u23f0 Dur\u00e9e : 15 min<\/p>\n\n\n\n

\ud83d\udcaa Difficult\u00e9 : niveau 1\/3<\/p>\n\n\n\n

\nD\u00e9terminer un \u00e9quivalent simple des suites d\u00e9finies par :<\/p>\n\n\n\n

\n
\n

<\/p>\n1. \"u_n=\sqrt{n+1}-\sqrt{n-1}\";\n\n\n\n

<\/p>\n3. \"u_n=a^n-b^n\" avec \"(a,b)\in\big(\mathbb{R}_+^*\big)^2\" et \"a\neq b\".\n<\/div>\n\n\n\n

\n

<\/p>\n2. \"\displaystyle v_n=\frac{\left(e^{\frac{1}{n}}-1\right) \sin\left(\frac{1}{n^3}\right)}{\tan\left(\frac{1}{n}\right)\left(\sqrt[3]{1+\frac{1}{n}}-1\right)}\";\n<\/div>\n<\/div>\n\n\n\n

Corrig\u00e9 de l’exercice<\/h2>\n\n\n\n

<\/p>\n1. En multipliant par la quantit\u00e9 conjugu\u00e9e, on a \"\displaystyle u_n=\frac{2}{\sqrt{1+n}+\sqrt{1-n}}=\frac{2}{\sqrt{n}\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}\right)}\".
\nOr, par continuit\u00e9 de la fonction \"\sqrt{\cdot}\", on a \"\displaystyle \sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}\xrightarrow[n\to+\infty]{}2\". Donc, \"\displaystyle u_n\sim\frac{1}{\sqrt{n}}\".
\n2. On a : \"e^u-1\underset{u\to 0}{\sim}u\", \"\sin(u)\underset{u\to 0}{\sim}u\", \"\tan(u)\underset{u\to 0}{\sim}u\" et \"\displaystyle\sqrt[3]{1+u}-1=(1+u)^{\frac{1}{3}}-1\underset{u\to 0}{\sim}\frac{u}{3}\".
\nOr, \"\displaystyle\frac{1}{n}\xrightarrow[n\to+\infty]{}0\".
\nDonc, par substitution, \"\displaystyle v_n \sim \frac{\frac{1}{n}\times \frac{1}{n^3} }{\frac{1}{n}\times \frac{1}{3n}}\sim \frac{3}{n^2}\".
\n3. Il y a deux cas.
\nCas 1 : \"0<a<b\". On a : \"\displaystyle u_n=b^n\left(\left(\frac{a}{b}\right)^n-1\right)\". Or, \"\displaystyle \left|\frac{a}{b}\right|<1\".
\nDonc, \"\displaystyle \left(\frac{a}{b}\right)^n=o(1)\".
\nDonc, \"u_n=b^n(-1+o(1))\", donc, \"u_n\sim -b^n\".
\nCas 2 : \"0<b<a\". On a : \"\displaystyle u_n=a^n\left(1-\left(\frac{b}{a}\right)^n\right)\". Or, \"\displaystyle \left|\frac{b}{a}\right|<1\".
\nDonc, \"\displaystyle \left(\frac{b}{a}\right)^n=o(1)\".
\nDonc, \"u_n=a^n(1+o(1))\", donc, \"u_n\sim a^n\".\n\n\n\n

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\"livre<\/a><\/figure>\n<\/div>\n\n\n\n
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Cet article est extrait de l’ouvrage Maths MPSI-MP2I. Tout-en-un : cours, m\u00e9thodes, entra\u00eenement et corrig\u00e9s <\/a><\/em>(\u00e9ditions Vuibert, juin 2021) \u00e9crit par E. Thomas, S. Bellec, G. Boutard. ISBN n\u00b09782311408720<\/em><\/p>\n\n <\/div>\n <\/section>\n<\/div>\n<\/div>\n\n\n

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\n Tu as aim\u00e9 cet article ?<\/span>\n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":"

Besoin d’aide sur un exercice portant sur l’analyse asymptotique ? Pas de panique ! Gr\u00e2ce \u00e0 ce cours (…)<\/p>\n","protected":false},"author":158,"featured_media":244610,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"category":[803,810],"tag":[78,345],"class_list":["post-238548","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-apprendre-matiere","category-maths","tag-prepa","tag-prepa-scientifique"],"acf":[],"_links":{"self":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/238548","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/users\/158"}],"replies":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/comments?post=238548"}],"version-history":[{"count":0,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/238548\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media\/244610"}],"wp:attachment":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media?parent=238548"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/category?post=238548"},{"taxonomy":"tag","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/tag?post=238548"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}