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let-f-x-e-x-1-x-sin-3x-1-dtermine-f-n-x-and-f-n-0-2-developp-f-at-integr-serie-




Question Number 61329 by maxmathsup by imad last updated on 01/Jun/19
let f(x) =(e^(−x) /(1+x)) sin(3x)  1) dtermine f^((n)) (x) and f^((n)) (0)  2) developp f at integr serie .
letf(x)=ex1+xsin(3x)1)dterminef(n)(x)andf(n)(0)2)developpfatintegrserie.
Commented by maxmathsup by imad last updated on 03/Jun/19
1)  we have f(x)=Im((1/(1+x))e^(−x+3ix) ) = Im((1/(1+x)) e^((−1+3i)x) )=Im(w(x)) we have  w(x)^((n)) ={(1/(1+x)) e^((−1+3i)x) }^((n))  =_(leibniz)       Σ_(k=0) ^n  C_n ^k  ((1/(1+x)))^((k))  (e^((−1+3i)x) )^((n−k))   but (e^(zx) )^((k))  =_(zfixed)     z^k  e^(zx)  ⇒{e^((−1+3i)x) }^((n−k)) =(−1+3i)^(n−k)  e^((−1+3i)x)    also  ((1/(1+x)))^((k))  =(((−1)^k  k!)/((1+x)^(k+1) )) ⇒w^((n)) (x) = Σ_(k=0) ^n  C_n ^k   (((−1)^k  k!)/((1+x)^(k+1) )) (−1+3i)^(n−k)  e^((−1+3i)x)     we have ∣−1+3i∣ =(√(1+9))=(√(10)) ⇒−1+3i =(√(10))(((−1)/( (√(10)))) +((3i)/( (√(10)))))=r e^(iθ)  ⇒  r=(√(10))   and   cosθ =((−1)/( (√(10)))) , sinθ =(3/( (√(10)))) ⇒tanθ =−3 ⇒θ =−arctan(3) ⇒  −1+3i =(√(10))e^(−i arctan(3))  =⇒(−1+3i)^(n−k)  =10^((n−k)/2)   e^(−(n−k)i arctan(3))   and  (−1+3i)^(n−k)  e^((−1+3i)x)  =  10^((n−k)/2)  (cos(n−k)arctan3)−isin(n−k)arctan(3)}e^(−x) {cos(3x)+isin(3x)}  e^(−x)  10^((n−k)/2) {cos(3x)cos(n−k)arctan3 +isin(3x)cos(n−k)arctan3  −i cos(3x) sin(n−k)arctan3 +sin(3x)sin(n−k)arctan3  ⇒ f^((n)) (x)=Im(w^((n)) )=  Σ_(k=0) ^n    (−1)^k k!  (C_n ^k /((x+1)^(k+1) )) e^(−x)  10^((n−k)/2) {  sin(3x)cos{(n−k)arctan(3)  −cos(3x) sin{(n−k)arctan(3)} ⇒  f^((n)) (0) =− Σ_(k=0) ^n   (−1)^k k!   C_n ^k   10^((n−k)/2)   sin{(n−k)arctan3}
1)wehavef(x)=Im(11+xex+3ix)=Im(11+xe(1+3i)x)=Im(w(x))wehavew(x)(n)={11+xe(1+3i)x}(n)=leibnizk=0nCnk(11+x)(k)(e(1+3i)x)(nk)but(ezx)(k)=zfixedzkezx{e(1+3i)x}(nk)=(1+3i)nke(1+3i)xalso(11+x)(k)=(1)kk!(1+x)k+1w(n)(x)=k=0nCnk(1)kk!(1+x)k+1(1+3i)nke(1+3i)xwehave1+3i=1+9=101+3i=10(110+3i10)=reiθr=10andcosθ=110,sinθ=310tanθ=3θ=arctan(3)1+3i=10eiarctan(3)=⇒(1+3i)nk=10nk2e(nk)iarctan(3)and(1+3i)nke(1+3i)x=10nk2(cos(nk)arctan3)isin(nk)arctan(3)}ex{cos(3x)+isin(3x)}ex10nk2{cos(3x)cos(nk)arctan3+isin(3x)cos(nk)arctan3icos(3x)sin(nk)arctan3+sin(3x)sin(nk)arctan3f(n)(x)=Im(w(n))=k=0n(1)kk!Cnk(x+1)k+1ex10nk2{sin(3x)cos{(nk)arctan(3)cos(3x)sin{(nk)arctan(3)}f(n)(0)=k=0n(1)kk!Cnk10nk2sin{(nk)arctan3}
Commented by maxmathsup by imad last updated on 03/Jun/19
2)  f(x) =Σ_(n=0) ^∞   ((f^((n)) (0))/(n!)) x^n   =Σ_(n=0) ^∞   (1/(n!)){ Σ_(k=0) ^n (−1)^(k+1)  k!   ((n!)/(k!(n−k)!)) 10^((n−k)/2)  sin{(n−k)arctan(3)}  =Σ_(n=0) ^∞   (Σ_(k=0) ^n  (−1)^(k+1)    ((10^((n−k)/2) )/((n−k)!)) sin{(n−k)arctan3})x^n  ⇒  f(x) =Σ_(n=0) ^∞  a_n x^n   with    a_n =Σ_(k=0) ^n  (−1)^(k+1)  ((10^((n−k)/2) )/((n−k)!)) sin{(n−k)arctan3}
2)f(x)=n=0f(n)(0)n!xn=n=01n!{k=0n(1)k+1k!n!k!(nk)!10nk2sin{(nk)arctan(3)}=n=0(k=0n(1)k+110nk2(nk)!sin{(nk)arctan3})xnf(x)=n=0anxnwithan=k=0n(1)k+110nk2(nk)!sin{(nk)arctan3}

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