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let-g-x-x-1-x-2-x-1-1-find-g-n-x-2-calculate-g-n-0-3-developp-g-at-integr-serie-

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Question Number 42688 by prof Abdo imad last updated on 31/Aug/18
let g(x) =((x−1)/(x^2 +x +1)) 1) find g^((n)) (x) 2)calculate g^((n)) (0) 3) developp g at integr serie.
letg(x)=x−1x2+x+11)findg(n)(x)2)calculateg(n)(0)3)developpgatintegrserie.
Commented by maxmathsup by imad last updated on 02/Sep/18
1) poles of g? roots of x^2 +x+1 are j=e^(i((2π)/3)) and j^− =e^(−((i2π)/3)) g(x)=((x−1)/((x−j)(x−j^− ))) =(a/(x−j)) +(b/(x−j^− )) a =lim_(x→j) (x−j)g(x) =((j−1)/(j−j^− )) =((j−1)/(2 i((√3)/2))) =((j−1)/(i(√3))) b =lim_(x→j^− ) (x−j^− )g(x) =((j^− −1)/((j^− −j))) =((j^− −1)/(−i(√3))) =((1−j^− )/(i(√3))) ⇒ g(x) = ((j−1)/(i(√3)(x−j))) −((j^− −1)/(i(√3)(x−j^− ))) ⇒ g^((n)) (x)=((j−1)/(i(√3))) (((−1)^n n!)/((x−j)^(n+1) )) −((j^− −1)/(i(√3))) (((−1)^n n!)/((x−j^− )^(n+1) )) =(((−1)^n n!)/(i(√3))){ ((j−1)/((x−j)^(n+1) )) −((j^− −1)/((x−j^− )^(n+1) ))} =(((−1)^n n!)/(i(√3))){ (((j−1)(x−j^− )^(n+1) −(j^− −1)(x−j)^(n+1) )/((x^2 +x+1)^(n+1) ))}
1)polesofg?rootsofx2+x+1arej=ei2π3andj−=e−i2π3g(x)=x−1(x−j)(x−j−)=ax−j+bx−j−a=limx→j(x−j)g(x)=j−1j−j−=j−12i32=j−1i3b=limx→j−(x−j−)g(x)=j−−1(j−−j)=j−−1−i3=1−j−i3⇒g(x)=j−1i3(x−j)−j−−1i3(x−j−)⇒g(n)(x)=j−1i3(−1)nn!(x−j)n+1−j−−1i3(−1)nn!(x−j−)n+1=(−1)nn!i3{j−1(x−j)n+1−j−−1(x−j−)n+1}=(−1)nn!i3{(j−1)(x−j−)n+1−(j−−1)(x−j)n+1(x2+x+1)n+1}
Commented by maxmathsup by imad last updated on 02/Sep/18
2) g^((n)) (0) = (((−1)^n n!)/(i(√3))){ (j−1)(−j^− )^(n+1) −(j^− −1)(−j)^(n+1) } =−((n!)/(i(√3))){ (j−1)(j^− )^(n+1) −(j^− −1)j^(n+1) } =((n!)/(i(√3))) 2iIm((j^− −1)(j)^(n+1) ) but (j^− −1)j^(n+1) =( e^(−((i2π)/3)) −1)e^(i((2(n+1)π)/3)) =(−(1/2) −i((√3)/2))(cos(((2(n+1)π)/3))+i sin(((2(n+1)π)/3))) =−(1/2) cos(((2(n+1)π)/3))−(i/2) sin(((2(n+1)π)/3))−((i(√3))/2) cos(((2(n+1)π)/3)) +((√3)/2) cos(((2(n+1)π)/3))sin(((2(n+1)π)/3)) ⇒ g^((n)) (0) =((2n!)/( (√3))) {−(1/2) sin(((2(n+1)π)/3))−((√3)/2) cos(((2(n+1)π)/3))} = −((n!)/( (√3))){ sin(((2(n+1)π)/3)) +(√3)cos(((2(n+1)π)/3)) .
2)g(n)(0)=(−1)nn!i3{(j−1)(−j−)n+1−(j−−1)(−j)n+1}=−n!i3{(j−1)(j−)n+1−(j−−1)jn+1}=n!i32iIm((j−−1)(j)n+1)but(j−−1)jn+1=(e−i2π3−1)ei2(n+1)π3=(−12−i32)(cos(2(n+1)π3)+isin(2(n+1)π3))=−12cos(2(n+1)π3)−i2sin(2(n+1)π3)−i32cos(2(n+1)π3)+32cos(2(n+1)π3)sin(2(n+1)π3)⇒g(n)(0)=2n!3{−12sin(2(n+1)π3)−32cos(2(n+1)π3)}=−n!3{sin(2(n+1)π3)+3cos(2(n+1)π3).
Commented by maxmathsup by imad last updated on 02/Sep/18
3) we have g(x) =Σ_(n=0) ^∞ ((g^((n)) (0))/(n!)) x^n ⇒ g(x) =−(1/( (√3))) Σ_(n=0) ^∞ { sin(((2(n+1)π)/3)) +(√3)cos(((2(n+1)π)/3))}x^n .
3)wehaveg(x)=∑n=0∞g(n)(0)n!xn⇒g(x)=−13∑n=0∞{sin(2(n+1)π3)+3cos(2(n+1)π3)}xn.

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