Testez vos connaissances sur la fonction digamma

Question 1 sur 10 10:00

[{"id":21798,"question":"Quelle est la relation entre la fonction digamma ψ(x) et la fonction gamma Γ(x) ?","option_a":"ψ(x) = Γ(x) \/ x","option_b":"ψ(x) = d\/dx [ln Γ(x)]","option_c":"ψ(x) = Γ'(x) \/ Γ(x)","option_d":"ψ(x) = ∫ Γ(t) dt","option_e":"","option_f":"","bonne_reponse":"B","explication":"La fonction digamma est définie comme la dérivée logarithmique de la fonction gamma : ψ(x) = d\/dx [ln Γ(x)] = Γ'(x) \/ Γ(x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":21799,"question":"La constante d'Euler-Mascheroni γ est égale à ψ(1).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"ψ(1) = -γ, donc γ = -ψ(1). La constante d'Euler-Mascheroni est positive, tandis que ψ(1) est négatif.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":21800,"question":"Quelle formule de récurrence est vérifiée par la fonction digamma ?","option_a":"ψ(x+1) = ψ(x) + 1\/x","option_b":"ψ(x+1) = ψ(x) - 1\/x","option_c":"ψ(x+1) = x ψ(x)","option_d":"ψ(x+1) = ψ(x) \/ x","option_e":"","option_f":"","bonne_reponse":"A","explication":"La fonction digamma satisfait la relation de récurrence : ψ(x+1) = ψ(x) + 1\/x.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":21801,"question":"Pour x = 2, quelle est la valeur de ψ(2) ?","option_a":"ψ(2) = -γ + 1","option_b":"ψ(2) = -γ - 1","option_c":"ψ(2) = γ","option_d":"ψ(2) = 1\/γ","option_e":"","option_f":"","bonne_reponse":"A","explication":"En utilisant la relation de récurrence ψ(x+1) = ψ(x) + 1\/x, on obtient ψ(2) = ψ(1) + 1 = -γ + 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":21802,"question":"La fonction digamma est-elle définie pour x = 0 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"La fonction digamma ψ(x) n'est pas définie pour x ≤ 0 car la fonction gamma Γ(x) a des pôles en ces points.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":21803,"question":"Quelle est la limite de ψ(x) lorsque x tend vers l'infini ?","option_a":"ψ(x) → 0","option_b":"ψ(x) → +∞","option_c":"ψ(x) → -∞","option_d":"ψ(x) → ln(x)","option_e":"","option_f":"","bonne_reponse":"D","explication":"Lorsque x tend vers l'infini, ψ(x) ~ ln(x) - 1\/(2x) - 1\/(12x²) + ..., donc ψ(x) → ln(x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":21804,"question":"La fonction digamma est-elle une fonction paire ou impaire ?","option_a":"Paire","option_b":"Impaire","option_c":"Ni paire ni impaire","option_d":"Les deux","option_e":"","option_f":"","bonne_reponse":"C","explication":"La fonction digamma n'est ni paire ni impaire. Par exemple, ψ(-1\/2) = -2 ln(2) - γ\/2 - ψ(1\/2), ce qui montre l'absence de symétrie.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":21805,"question":"Quelle intégrale est liée à la fonction digamma ?","option_a":"∫₀^∞ e^(-t) ln(t) dt = -γ","option_b":"∫₀^∞ e^(-t) t dt = ψ(1)","option_c":"∫₀^∞ e^(-t) \/ t dt = ψ(0)","option_d":"∫₀^∞ e^(-t) ln(t) dt = ψ(1)","option_e":"","option_f":"","bonne_reponse":"A","explication":"L'intégrale ∫₀^∞ e^(-t) ln(t) dt = -γ, où γ est la constante d'Euler-Mascheroni, est directement liée à la fonction digamma.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":21806,"question":"La fonction digamma peut-elle être exprimée comme une série ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Oui, la fonction digamma admet un développement en série : ψ(x) = -γ + ∑ (1\/k - 1\/(x+k)) pour k de 1 à ∞.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":21807,"question":"Quelle est la valeur de ψ(1\/2) ?","option_a":"ψ(1\/2) = -γ - 2 ln(2)","option_b":"ψ(1\/2) = -γ + 2 ln(2)","option_c":"ψ(1\/2) = γ","option_d":"ψ(1\/2) = 2 ln(2)","option_e":"","option_f":"","bonne_reponse":"A","explication":"La valeur de ψ(1\/2) est connue : ψ(1\/2) = -γ - 2 ln(2). Cette relation est utile dans de nombreux calculs intégrals.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]