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Question Number 35987 by abdo mathsup 649 cc last updated on 26/May/18
let f(x) = (1/(1+x^3 )) 1) calculate f^((n)) (x) 2) developp f at integr serie.
letf(x)=11+x31)calculatef(n)(x)2)developpfatintegrserie.
Commented by prof Abdo imad last updated on 27/May/18
we have f(x)= (1/((x+1)(x^2 −x+1))) let decompose f inside C[x] roots of x^2 −x +1 Δ=1−4=−3=(i(√3))^2 ⇒x_1 =((1 +i(√3))/2) x_2 =((1−(√3))/2) or j =−(1/2) +i((√3)/2) ⇒x_2 =−j and x_1 =−j^− ⇒f(x)=(1/((x+1)(x+j)(x+j^− ))) =(a/(x+1)) +(b/(x+j)) +(c/(x+j^− )) a =lim_(x→−1) (x+1)f(x)= (1/3) b = lim_(x→−j) (x+j)f(x) = (1/((−j+1)(−j+j^− ))) = (1/((j−1)(j−j^− ))) =(1/( (√3)(−(3/2)+i((√3)/2)))) = (2/(3( −(√3) +i))) = ((2( −(√3)−i))/(3(3 +1))) = ((−(√3)−i)/6) c =lim_(x→−j^− ) (x+j^− )f(x) = (1/((−j^− +1)(j−j^− ))) = (1/( (√3)( ((1+i(√3))/2) +1))) = (1/( (√3)( (3/2) +i((√3)/2)))) = (2/(3((√3) +i))) =((2((√3)−i))/(3(4))) = (((√3)−i)/6) ⇒ f(x) = (1/(3(x+1))) −(((√3)+i)/(6(x+j))) +(((√3)−i)/(6(x+j^− ))) ⇒ f^((n)) (x) = (1/3) (((−1)^n n!)/((x+1)^(n+1) )) −(((√3)+i)/6) (((−1)^n n!)/((x+j)^(n+1) )) + (((√3)−i)/6) (((−1)^n n!)/((x+j^− )^(n+1) ))
wehavef(x)=1(x+1)(x2−x+1)letdecomposefinsideC[x]rootsofx2−x+1Δ=1−4=−3=(i3)2⇒x1=1+i32x2=1−32orj=−12+i32⇒x2=−jandx1=−j−⇒f(x)=1(x+1)(x+j)(x+j−)=ax+1+bx+j+cx+j−a=limx→−1(x+1)f(x)=13b=limx→−j(x+j)f(x)=1(−j+1)(−j+j−)=1(j−1)(j−j−)=13(−32+i32)=23(−3+i)=2(−3−i)3(3+1)=−3−i6c=limx→−j−(x+j−)f(x)=1(−j−+1)(j−j−)=13(1+i32+1)=13(32+i32)=23(3+i)=2(3−i)3(4)=3−i6⇒f(x)=13(x+1)−3+i6(x+j)+3−i6(x+j−)⇒f(n)(x)=13(−1)nn!(x+1)n+1−3+i6(−1)nn!(x+j)n+1+3−i6(−1)nn!(x+j−)n+1
Commented by prof Abdo imad last updated on 27/May/18
f^n (x) = (((−1)^n n!)/(3(x+1)^(n+1) )) + (((−1)^n n!)/6){ (((√3)−i)/((x+j^− )^(n+1) )) −(((√3)+i)/((x+j)^(n+1) ))} f^((n)) (x) = (((−1)^n n!)/(3(x+1)^(n+1) )) +(((−1)^n n!)/6){ ((((√3)−i)/((x+j^− )^(n+1) ))) −conj( (((√3)−i)/((x+j^− )^(n+1) ))) = (((−1)^n n!)/(3(x+1)^(n+1) )) +(((−1)^n n!)/3) Im( (((√3)−i)/((x +j^− )^(n+1) )))
fn(x)=(−1)nn!3(x+1)n+1+(−1)nn!6{3−i(x+j−)n+1−3+i(x+j)n+1}f(n)(x)=(−1)nn!3(x+1)n+1+(−1)nn!6{(3−i(x+j−)n+1)−conj(3−i(x+j−)n+1)=(−1)nn!3(x+1)n+1+(−1)nn!3Im(3−i(x+j−)n+1)
Commented by prof Abdo imad last updated on 27/May/18
2) we have f(x) = Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n but f^((n)) (0) = (((−1)^n n!)/3) −(((√3)+i)/6) (((−1)^n n!)/j^(n+1) ) +(((√3)−i)/6) (((−1)^n n!)/j^−^(n+1) ) =(((−1)^n n!)/3) + (((−1)^n n!)/6){ (((√3)−i)/j^−^(n+1) ) −(((√3)+i)/j^(n+1) )} =(((−1)^n n!)/3) +(((−1)^n n!)/6){((((√3)−i)j^(n+1) −((√3)+i)j^−^(n+1) )/1)} = (((−1)^n n!)/3) +(((−1)^n n!)/3) Im{ ((√3)−i)j^(n+1) } but ((√3)−i)j^(n+1) =2 (((√3)/2)−(1/2))(e^(i((2π)/3)) )^(n+1) = 2 e^(−i(π/6)) e^(i(2/3)(n+1)π) = 2 e^(i( ((2(n+1)π)/3)−(π/6))) ⇒
2)wehavef(x)=∑n=0∞f(n)(0)n!xnbutf(n)(0)=(−1)nn!3−3+i6(−1)nn!jn+1+3−i6(−1)nn!j−n+1=(−1)nn!3+(−1)nn!6{3−ij−n+1−3+ijn+1}=(−1)nn!3+(−1)nn!6{(3−i)jn+1−(3+i)j−n+11}=(−1)nn!3+(−1)nn!3Im{(3−i)jn+1}but(3−i)jn+1=2(32−12)(ei2π3)n+1=2e−iπ6ei23(n+1)π=2ei(2(n+1)π3−π6)⇒
Commented by abdo.msup.com last updated on 27/May/18
f^((n)) (0) = (((−1)^n n!)/3) +(2/3)(−1)^n n!sin(((2(n+1)π)/3)−(π/6)) ⇒f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!))x^n =Σ_(n=0) ^∞ (((−1)^n )/3){ 1 +2sin(((2(n+1)π)/3) −(π/6))}x^n another method we have f(x) =(1/(1+x^3 )) so for ∣x∣<1 f(x) =Σ_(n=0) ^∞ (−1)^n x^(3n) .
f(n)(0)=(−1)nn!3+23(−1)nn!sin(2(n+1)π3−π6)⇒f(x)=∑n=0∞f(n)(0)n!xn=∑n=0∞(−1)n3{1+2sin(2(n+1)π3−π6)}xnanothermethodwehavef(x)=11+x3sofor∣x∣<1f(x)=∑n=0∞(−1)nx3n.

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