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Question Number 42999 by abdo.msup.com last updated on 06/Sep/18
letf(x)=x+1x−11)calculatef(n)(2)2)iff(x)=∑n=0∞an(x−2)nfindthesequencean
Commented by maxmathsup by imad last updated on 08/Sep/18
1)firstletfindf(n)(x)wehavef(n)(x)=(x)n+(1x−1)nbut{1x−1}(n)=(−1)nn!(x−1)n+1letdetermine(x)n={x12}(n)letαfromQ{xα}(1)=αxα−1⇒{xα}(2)=α(α−1)xα−2⇒{xα}(n)=α(α−1)...(α−n+1)xα−nso{x12}(n)=12(12−1)(12−2)....(12−n+1)x12−n=12(−12)(−32)....(1−2n+2)2)x12−n=12(−12)(−32)....(−2n−32)x12−n⇒f(n)(x)=12(−12)(−32)=(−2n−32)x12−n+(−1)nn!(x−1)n+1.⇒f(n)(2)=12(−12)(−32).....(−2n−32)212−n+(−1)nn!=12n2(−12)(−32)...(−2n−32)+(−1)nn!.2)developpementoff(x)atintegrserieserisatv(2)givef(x)=∑n=0∞f(n)(2)n!(x−2)n⇒an=f(n)(2)n!⇒an=1(n!)2n2(−12)(−32)...(−2n−32)+(−1)n.nt
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