Quiz — 4-TD suites et séries de fonctions.PDF

Question 1 sur 10 20:00

[{"id":58839,"question":"Soit la suite de fonctions (fn) définie par fn(x) = x^n sur [0,1]. Que peut-on dire de sa convergence ?","option_a":"La suite converge simplement vers une fonction discontinue en x=1","option_b":"La suite converge uniformément vers la fonction nulle","option_c":"La convergence est uniforme sur [0,1\/2] mais pas sur [0,1]","option_d":"La suite ne converge pas","option_e":"","option_f":"","bonne_reponse":"c","explication":"La suite (fn) converge simplement vers la fonction nulle sur [0,1], mais la convergence n'est pas uniforme sur cet intervalle car fn(1) = 1 pour tout n. Cependant, elle est uniforme sur [0,1\/2] car sup{|fn(x)| : x ∈ [0,1\/2]} = (1\/2)^n → 0.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"c\", \"options\": {\"a\": \"La suite converge simplement vers une fonction discontinue en x=1","_debug_options_count":4},{"id":58840,"question":"La série entière Σ (x^n \/ n!) converge-t-elle uniformément sur ℝ ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"Faux. La série converge simplement sur ℝ (rayon de convergence infini), mais pas uniformément car la série des normes (Σ |x^n \/ n!|) diverge pour x ≠ 0.","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":58841,"question":"Soit la série de fonctions Σ (sin(nx) \/ n^2). Quel critère peut-on utiliser pour étudier sa convergence uniforme ?","option_a":"Critère de Cauchy","option_b":"Critère de Weierstrass","option_c":"Critère de d'Alembert","option_d":"Critère de Leibniz","option_e":"","option_f":"","bonne_reponse":"b","explication":"Le critère de Weierstrass est adapté ici car |sin(nx)\/n^2| ≤ 1\/n^2, et la série Σ 1\/n^2 converge (série de Riemann).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Critère de Cauchy\", \"b\": \"Critère de Weierstrass\", \"c\": \"Critè","_debug_options_count":4},{"id":58842,"question":"La série entière Σ (n x^n) a pour rayon de convergence :","option_a":"R = 0","option_b":"R = 1","option_c":"R = +∞","option_d":"R = 1\/2","option_e":"","option_f":"","bonne_reponse":"b","explication":"En appliquant le critère de d'Alembert, on trouve lim |(n+1)\/n| = 1, donc R = 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"R = 0\", \"b\": \"R = 1\", \"c\": \"R = +∞\", \"d\": \"R = 1\/2\"}}","_debug_options_count":4},{"id":58843,"question":"Peut-on dériver terme à terme la série Σ (x^n \/ n) sur ]-1,1[ ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Vrai. La série des dérivées Σ (x^(n-1)) converge uniformément sur [-a,a] pour tout a \u003C 1, donc on peut dériver terme à terme.","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":58844,"question":"Soit f(x) = e^x. Quel est le développement limité de f à l'ordre 3 autour de 0 ?","option_a":"1 + x + x^2\/2 + x^3\/6","option_b":"1 + x + x^2 + x^3","option_c":"1 + x^2\/2 + x^3\/6","option_d":"x + x^2\/2 + x^3\/6","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le développement limité de e^x à l'ordre 3 est 1 + x + x^2\/2! + x^3\/3! = 1 + x + x^2\/2 + x^3\/6.","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 + x^2\/2 + x^3\/6\", \"b\": \"1 + x + x^2 + x^3\", \"c\": \"1 + x^2\/2","_debug_options_count":4},{"id":58845,"question":"La série Σ (1\/n^2) est-elle normalement convergente sur ℕ* ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Vrai. La série Σ |1\/n^2| = Σ 1\/n^2 converge (série de Riemann), donc la série est normalement convergente.","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":58846,"question":"Soit la série entière Σ (x^n \/ (n+1)). Quel est son rayon de convergence ?","option_a":"R = 0","option_b":"R = 1","option_c":"R = +∞","option_d":"R = 2","option_e":"","option_f":"","bonne_reponse":"b","explication":"En appliquant le critère de d'Alembert, on trouve lim |(n+2)\/(n+1)| = 1, donc R = 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"R = 0\", \"b\": \"R = 1\", \"c\": \"R = +∞\", \"d\": \"R = 2\"}}","_debug_options_count":4},{"id":58847,"question":"La fonction f(x) = 1\/(1-x) admet pour développement en série entière autour de 0 :","option_a":"Σ x^n pour |x| \u003C 1","option_b":"Σ (-1)^n x^n pour |x| \u003C 1","option_c":"Σ x^(2n) pour |x| \u003C 1","option_d":"Σ n x^n pour |x| \u003C 1","option_e":"","option_f":"","bonne_reponse":"a","explication":"La série géométrique donne 1\/(1-x) = Σ x^n pour |x| \u003C 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Σ x^n pour |x| \u003C 1\", \"b\": \"Σ (-1)^n x^n pour |x| \u003C 1\", \"c\": \"Σ","_debug_options_count":4},{"id":58848,"question":"Soit la série de fonctions Σ (x^n \/ (1 + n^2)). Peut-on appliquer le théorème d'intégration terme à terme sur [0,1] ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Vrai. La série converge uniformément sur [0,1] (critère de Weierstrass avec 1\/(1+n^2) ≤ 1\/n^2), donc on peut intégrer terme à terme.","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}]