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let-f-x-x-x-2-x-1-1-calculate-f-n-0-2-developp-f-at-integr-serie-




Question Number 36010 by abdo mathsup 649 cc last updated on 27/May/18
let f(x)=  (x/(x^2  +x+1))  1) calculate f^((n)) (0)  2) developp f  at integr serie .
letf(x)=xx2+x+11)calculatef(n)(0)2)developpfatintegrserie.
Commented by prof Abdo imad last updated on 28/May/18
let decompose f(x)inside C(x) the roots of  p(x)=x^2  +x+1 are are j =e^(i((2π)/3))  and j^−  =e^(−i((2π)/3))  so  f(x)=  (x/((x−j)(x−j^− ))) = (a/(x−j)) +(b/(x−j^− ))  a=lim_(x→j) (x−j)f(x) = (j/(j−j^− )) = (j/(2 ((√3)/2))) = (j/( (√3)))  b =lim_(x→j^− ) (x−j^− )f(x)= (j^− /(j^− −j)) =−(j^− /( (√3))) ⇒  f(x) = (j/( (√3)(x−j))) −(j^− /( (√3)(x−j^− ))) ⇒  f^((n)) (x) = (j/( (√3))){ (((−1)^n n!)/((x−j)^(n+1) ))} −(j^− /( (√3))) { (((−1)^n  n!)/((x−j^− )^(n+1) ))}  f^((n)) (x) = (((−1)^n n!)/( (√3))){  (j/((x−j)^(n+1) )) −(j^− /((x−j^− )^(n+1) ))}  =(((−1)^n n!)/( (√3))){ ((j(x−j^− )^(n+1)  −j^− (x−j)^(n+1) )/((x^2  +x+1)^(n+1) ))}
letdecomposef(x)insideC(x)therootsofp(x)=x2+x+1arearej=ei2π3andj=ei2π3sof(x)=x(xj)(xj)=axj+bxja=limxj(xj)f(x)=jjj=j232=j3b=limxj(xj)f(x)=jjj=j3f(x)=j3(xj)j3(xj)f(n)(x)=j3{(1)nn!(xj)n+1}j3{(1)nn!(xj)n+1}f(n)(x)=(1)nn!3{j(xj)n+1j(xj)n+1}=(1)nn!3{j(xj)n+1j(xj)n+1(x2+x+1)n+1}
Commented by prof Abdo imad last updated on 28/May/18
so f^((n)) (0) = (((−1)^n n!)/( (√3))){ ((j(−j^− )^(n+1)  −j^− (−j)^(n+1) )/1)}  =(((−1)^n n!)/( (√3))){  (−1)^(n+1)  j(j^− )^(n+1)  −(−1)^(n+1) j^−  j^(n+1) }  =(((−1)^n n!)/( (√3))) (−1)^(n+1) {  j^−^n    −j^n }  = (1/( (√3))){  2Im(j^n )) = (2/( (√3))) sin(((2nπ)/3))  f^((n)) (0) = (2/( (√3)))sin(((2nπ)/3)).  2) we have f(x) = Σ_(n=0) ^∞   ((f^((n)) (0))/(n!)) x^n  ⇒  f(x) = Σ_(n=0) ^∞     (2/( (√3)))  ((sin(((2nπ)/3)))/(n!)) x^n   .  f(x) = (2/( (√3))) Σ_(n=0) ^∞    ((sin(((2nπ)/3)))/(n!)) x^n   .
sof(n)(0)=(1)nn!3{j(j)n+1j(j)n+11}=(1)nn!3{(1)n+1j(j)n+1(1)n+1jjn+1}=(1)nn!3(1)n+1{jnjn}=13{2Im(jn))=23sin(2nπ3)f(n)(0)=23sin(2nπ3).2)wehavef(x)=n=0f(n)(0)n!xnf(x)=n=023sin(2nπ3)n!xn.f(x)=23n=0sin(2nπ3)n!xn.

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