Quiz — taylor.pdf

Question 1 sur 10 20:00

[{"id":114734,"question":"Quel est le développement limité de \u003Cem\u003Ee^x\u003C\/em\u003E à l'ordre 3 autour de 0 ?","option_a":"1 + x + x²\/2 + x³\/6","option_b":"1 + x + x²\/2 + x³\/3","option_c":"x + x²\/2 + x³\/6","option_d":"1 + x + x² + x³","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le développement limité de \u003Cem\u003Ee^x\u003C\/em\u003E à l'ordre 3 est donné par la formule de Taylor : \u003Cem\u003Ee^x ≈ 1 + x + x²\/2! + x³\/3!\u003C\/em\u003E.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"1 + x + x²\/2 + x³\/6\", \"b\": \"1 + x + x²\/2 + x³\/3\", \"c\": \"x + x","_debug_options_count":4},{"id":114735,"question":"La série de Taylor d'une fonction \u003Cem\u003Ef\u003C\/em\u003E autour de \u003Cem\u003Ea\u003C\/em\u003E converge toujours vers \u003Cem\u003Ef(x)\u003C\/em\u003E pour tout \u003Cem\u003Ex\u003C\/em\u003E.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"La convergence n'est pas systématique. Par exemple, la série de Taylor de \u003Cem\u003Ef(x) = e^(-1\/x²)\u003C\/em\u003E autour de 0 ne converge pas vers \u003Cem\u003Ef(0)\u003C\/em\u003E.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":114736,"question":"Quel est le reste de Lagrange pour un développement limité d'ordre \u003Cem\u003En\u003C\/em\u003E de \u003Cem\u003Ef\u003C\/em\u003E autour de \u003Cem\u003Ea\u003C\/em\u003E ?","option_a":"\u003Cem\u003ER_n(x) = f^(n+1)(c) * (x-a)^(n+1) \/ (n+1)!\u003C\/em\u003E","option_b":"\u003Cem\u003ER_n(x) = f^(n)(c) * (x-a)^n \/ n!\u003C\/em\u003E","option_c":"\u003Cem\u003ER_n(x) = f^(n+1)(c) * (x-a)^n \/ n!\u003C\/em\u003E","option_d":"\u003Cem\u003ER_n(x) = f^(n)(c) * (x-a)^(n+1) \/ (n+1)!\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le reste de Lagrange s'exprime par \u003Cem\u003ER_n(x) = f^(n+1)(c) * (x-a)^(n+1) \/ (n+1)!\u003C\/em\u003E où \u003Cem\u003Ec\u003C\/em\u003E est un point entre \u003Cem\u003Ea\u003C\/em\u003E et \u003Cem\u003Ex\u003C\/em\u003E.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"\u003Cem\u003ER_n(x) = f^(n+1)(c) * (x-a)^(n+1) \/ (n+1)!\u003C\/em\u003E\", \"b\": \"\u003Cem\u003ER","_debug_options_count":4},{"id":114737,"question":"Pour la fonction \u003Cem\u003Ef(x) = ln(1+x)\u003C\/em\u003E, quel est le développement limité à l'ordre 2 autour de 0 ?","option_a":"\u003Cem\u003Ex - x²\/2\u003C\/em\u003E","option_b":"\u003Cem\u003Ex + x²\/2\u003C\/em\u003E","option_c":"\u003Cem\u003E1 + x - x²\/2\u003C\/em\u003E","option_d":"\u003Cem\u003Ex - x²\/3\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"a","explication":"La dérivée première de \u003Cem\u003Eln(1+x)\u003C\/em\u003E est \u003Cem\u003E1\/(1+x)\u003C\/em\u003E, et la dérivée seconde est \u003Cem\u003E-1\/(1+x)²\u003C\/em\u003E. Ainsi, \u003Cem\u003Eln(1+x) ≈ x - x²\/2\u003C\/em\u003E à l'ordre 2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"\u003Cem\u003Ex - x²\/2\u003C\/em\u003E\", \"b\": \"\u003Cem\u003Ex + x²\/2\u003C\/em\u003E\", \"c\": \"\u003Cem\u003E1 + x -","_debug_options_count":4},{"id":114738,"question":"Le développement limité de \u003Cem\u003Esin(x)\u003C\/em\u003E à l'ordre 5 autour de 0 est toujours exact pour tout \u003Cem\u003Ex\u003C\/em\u003E.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"Le développement limité de \u003Cem\u003Esin(x)\u003C\/em\u003E est une série infinie. L'approximation à l'ordre 5 n'est exacte que pour \u003Cem\u003Ex\u003C\/em\u003E proche de 0, avec une erreur donnée par le reste de Lagrange.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":114739,"question":"Quelle est la formule de Taylor-Young pour une fonction \u003Cem\u003Ef\u003C\/em\u003E dérivable \u003Cem\u003En\u003C\/em\u003E fois en \u003Cem\u003Ea\u003C\/em\u003E ?","option_a":"\u003Cem\u003Ef(x) = f(a) + f'(a)(x-a) + ... + f^(n)(a)(x-a)^n \/ n! + o((x-a)^n)\u003C\/em\u003E","option_b":"\u003Cem\u003Ef(x) = f(a) + f'(a)(x-a) + ... + f^(n)(a)(x-a)^n + o((x-a)^n)\u003C\/em\u003E","option_c":"\u003Cem\u003Ef(x) = f(a) + f'(a)(x-a) + ... + f^(n)(a)(x-a)^n \/ n!\u003C\/em\u003E","option_d":"\u003Cem\u003Ef(x) = f(a) + f'(a)(x-a) + ... + f^(n)(a)(x-a)^n \/ (n+1)!\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"a","explication":"La formule de Taylor-Young s'écrit \u003Cem\u003Ef(x) = f(a) + f'(a)(x-a) + ... + f^(n)(a)(x-a)^n \/ n! + o((x-a)^n)\u003C\/em\u003E.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"\u003Cem\u003Ef(x) = f(a) + f'(a)(x-a) + ... + f^(n)(a)(x-a)^n \/ n! + o((x-","_debug_options_count":4},{"id":114740,"question":"Si \u003Cem\u003Ef(x) = cos(x)\u003C\/em\u003E, quel est le développement limité à l'ordre 4 autour de 0 ?","option_a":"\u003Cem\u003E1 - x²\/2 + x⁴\/24\u003C\/em\u003E","option_b":"\u003Cem\u003E1 - x²\/2 + x⁴\/4\u003C\/em\u003E","option_c":"\u003Cem\u003Ex - x³\/6\u003C\/em\u003E","option_d":"\u003Cem\u003E1 - x + x²\/2 - x³\/6 + x⁴\/24\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"a","explication":"Les dérivées successives de \u003Cem\u003Ecos(x)\u003C\/em\u003E alternent entre \u003Cem\u003Ecos(x)\u003C\/em\u003E et \u003Cem\u003E-sin(x)\u003C\/em\u003E. Ainsi, \u003Cem\u003Ecos(x) ≈ 1 - x²\/2 + x⁴\/24\u003C\/em\u003E à l'ordre 4.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"\u003Cem\u003E1 - x²\/2 + x⁴\/24\u003C\/em\u003E\", \"b\": \"\u003Cem\u003E1 - x²\/2 + x⁴\/4\u003C\/em\u003E\"","_debug_options_count":4},{"id":114741,"question":"Le reste de Lagrange permet de majorer l'erreur d'approximation d'un développement limité.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le reste de Lagrange donne une expression de l'erreur \u003Cem\u003ER_n(x)\u003C\/em\u003E, ce qui permet de la majorer en utilisant des inégalités sur les dérivées.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":114742,"question":"Pour la fonction \u003Cem\u003Ef(x) = (1+x)^α\u003C\/em\u003E, quel est le développement limité à l'ordre 2 autour de 0 ?","option_a":"\u003Cem\u003E1 + αx + α(α-1)x²\/2\u003C\/em\u003E","option_b":"\u003Cem\u003E1 + αx + αx²\/2\u003C\/em\u003E","option_c":"\u003Cem\u003E1 + x + x²\/2\u003C\/em\u003E","option_d":"\u003Cem\u003Eαx + α(α-1)x²\/2\u003C\/em\u003E","option_e":"","option_f":"","bonne_reponse":"a","explication":"En utilisant la formule du binôme généralisé, \u003Cem\u003E(1+x)^α ≈ 1 + αx + α(α-1)x²\/2\u003C\/em\u003E à l'ordre 2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"\u003Cem\u003E1 + αx + α(α-1)x²\/2\u003C\/em\u003E\", \"b\": \"\u003Cem\u003E1 + αx + αx²\/2\u003C\/e","_debug_options_count":4},{"id":114743,"question":"La série de Maclaurin est un cas particulier de la série de Taylor où \u003Cem\u003Ea = 0\u003C\/em\u003E.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"La série de Maclaurin est bien définie par \u003Cem\u003Ef(x) = Σ f^(n)(0) * x^n \/ n!\u003C\/em\u003E, ce qui correspond à une série de Taylor centrée en \u003Cem\u003Ea = 0\u003C\/em\u003E.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4}]