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let-f-x-x-1-x-1-1-calculate-f-n-2-2-if-f-x-n-0-a-n-x-2-n-find-the-sequence-a-n-




Question Number 42999 by abdo.msup.com last updated on 06/Sep/18
let f(x)=(√x)+(1/(x−1))  1) calculate f^((n)) (2)  2) if f(x) =Σ_(n=0) ^∞  a_n (x−2)^n  find the  sequence a_n
letf(x)=x+1x11)calculatef(n)(2)2)iff(x)=n=0an(x2)nfindthesequencean
Commented by maxmathsup by imad last updated on 08/Sep/18
1) first let find f^((n)) (x)  we have f^((n)) (x)=((√x))^n  +((1/(x−1)))^n   but  {(1/(x−1))}^((n))  = (((−1)^n n!)/((x−1)^(n+1) )) let determine ((√x))^n  ={x^(1/2) }^((n)) let α fromQ  {x^α }^((1))  =α x^(α−1)  ⇒{x^α }^((2)) =α(α−1)x^(α−2)  ⇒{x^α }^((n))  =α(α−1)...(α−n+1)x^(α−n)   so { x^(1/2) }^((n)) =(1/2)((1/2)−1)((1/2)−2)....((1/2) −n+1)x^((1/2)−n)   =(1/2)(−(1/2))(−(3/2))....(((1−2n+2))/2))x^((1/2)−n)   =(1/2)(−(1/2))(−(3/2))....(−((2n−3)/2)) x^((1/2)−n)  ⇒  f^((n)) (x)=(1/2)(−(1/2))(−(3/2))=(−((2n−3)/2))x^((1/2)−n)  + (((−1)^n n!)/((x−1)^(n+1) )) .⇒  f^((n)) (2) =(1/2)(−(1/2))(−(3/2)).....(−((2n−3)/2))2^((1/2)−n)   +(−1)^n n!  = (1/(2^n (√2))) (−(1/2))(−(3/2))...(−((2n−3)/2)) +(−1)^n n! .  2) developpement of f(x) at integr serie seris at v(2) give  f(x) =Σ_(n=0) ^∞   ((f^((n)) (2))/(n!)) (x−2)^n  ⇒a_n =((f^((n)) (2))/(n!)) ⇒  a_n =  (1/((n!)2^n (√2)))(−(1/2))(−(3/2))...(−((2n−3)/2)) +(−1)^n  .    nt
1)firstletfindf(n)(x)wehavef(n)(x)=(x)n+(1x1)nbut{1x1}(n)=(1)nn!(x1)n+1letdetermine(x)n={x12}(n)letαfromQ{xα}(1)=αxα1{xα}(2)=α(α1)xα2{xα}(n)=α(α1)(αn+1)xαnso{x12}(n)=12(121)(122).(12n+1)x12n=12(12)(32).(12n+2)2)x12n=12(12)(32).(2n32)x12nf(n)(x)=12(12)(32)=(2n32)x12n+(1)nn!(x1)n+1.f(n)(2)=12(12)(32)..(2n32)212n+(1)nn!=12n2(12)(32)(2n32)+(1)nn!.2)developpementoff(x)atintegrserieserisatv(2)givef(x)=n=0f(n)(2)n!(x2)nan=f(n)(2)n!an=1(n!)2n2(12)(32)(2n32)+(1)n.nt

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