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let-f-x-x-x-3-2x-1-1-find-D-f-2-find-f-n-x-then-f-n-0-3-developp-f-at-integr-serie-

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Question Number 42689 by prof Abdo imad last updated on 31/Aug/18
let f(x) = (x/(x^3 −2x +1)) 1) find D_f 2) find f^((n)) (x) then f^((n)) (0) 3) developp f at integr serie.
letf(x)=xx3−2x+11)findDf2)findf(n)(x)thenf(n)(0)3)developpfatintegrserie.
Commented by maxmathsup by imad last updated on 01/Sep/18
1) x∈ D_f ⇔ x^3 −2x +1 ≠0 but x^3 −2x +1 =x^3 −1 −2x +2 =(x−1)(x^2 +x+1)−2(x−1) =(x−1)(x^2 +x+1 −2) =(x−1)(x^2 +x−1) roots of x^2 +x−1? Δ=1−4(−1)=5 ⇒α =((−1+(√5))/2) and β =((−1−(√5))/2) ⇒ D_f =R−{1,((−1+(√5))/2) ,((−1−(√5))/2)}
1)x∈Df⇔x3−2x+1≠0butx3−2x+1=x3−1−2x+2=(x−1)(x2+x+1)−2(x−1)=(x−1)(x2+x+1−2)=(x−1)(x2+x−1)rootsofx2+x−1?Δ=1−4(−1)=5⇒α=−1+52andβ=−1−52⇒Df=R−{1,−1+52,−1−52}
Commented by maxmathsup by imad last updated on 01/Sep/18
2) we have f(x)= (x/((x−1)(x−α)(x−β))) =(a/(x−1)) +(b/(x−α)) +(c/(x−β)) a =lim_(x→1) (x−1)f(x) = (1/((1−α)(1−β))) b =lim_(x→α) (x−α)f(x) = (α/((α−1)(α−β))) c =lim_(x→β) (x−β)f(x) = (β/((β−1)(β−α))) but α =((−1+(√5))/2) and β=((−1−(√5))/2) ⇒ a = (1/((1−((−1+(√5))/2))(1−((−1−(√5))/2)))) = (4/((3−(√5))(3+(√5)))) =(4/4) =1 b = ((−1+(√5))/(2(1−((−1+(√5))/2))(−(√5)))) =((−1+(√5))/( (√5)(3−(√5)))) =((−1+(√5))/(3(√5)−5)) . c = ((−1−(√5))/(2(1−((−1−(√5))/2))((√5)))) =((−1−(√5))/( (√5)(3−(√5)))) =((−1−(√5))/(3(√5)−5)) ⇒ f(x)= (1/(x−1)) +(1/(3(√5)−5)){ ((−1+(√5))/(x−α)) −((1+(√5))/(x−β))} ⇒ f^((n)) (x) = (((−1)^n n!)/((x−1)^(n+1) )) + ((−1+(√5))/(3(√5)−5)) (((−1)^n n!)/((x−α)^(n+1) )) −((1+(√5))/(3(√5)−5)) (((−1)^n n!)/((x−β)^(n+1) )) ⇒ f^((n)) (0) =−n! +((−1+(√5))/(3(√5)−5)) (((−1)^n n!)/((−α)^(n+1) )) −((1+(√5))/(3(√5)−5)) (((−1)^n n!)/((−β)^(n+1) )) =−n! −((−1+(√5))/(3(√5)−5)) ((n!)/α^(n+1) ) +((1+(√5))/(3(√5)−5)) ((n!)/β^(n+1) ) ⇒ f^((n)) (0) =−n! +((n!)/(3(√5)−5)){ ((1+(√5))/β^(n+1) ) +((1−(√5))/α^(n+1) )}
2)wehavef(x)=x(x−1)(x−α)(x−β)=ax−1+bx−α+cx−βa=limx→1(x−1)f(x)=1(1−α)(1−β)b=limx→α(x−α)f(x)=α(α−1)(α−β)c=limx→β(x−β)f(x)=β(β−1)(β−α)butα=−1+52andβ=−1−52⇒a=1(1−−1+52)(1−−1−52)=4(3−5)(3+5)=44=1b=−1+52(1−−1+52)(−5)=−1+55(3−5)=−1+535−5.c=−1−52(1−−1−52)(5)=−1−55(3−5)=−1−535−5⇒f(x)=1x−1+135−5{−1+5x−α−1+5x−β}⇒f(n)(x)=(−1)nn!(x−1)n+1+−1+535−5(−1)nn!(x−α)n+1−1+535−5(−1)nn!(x−β)n+1⇒f(n)(0)=−n!+−1+535−5(−1)nn!(−α)n+1−1+535−5(−1)nn!(−β)n+1=−n!−−1+535−5n!αn+1+1+535−5n!βn+1⇒f(n)(0)=−n!+n!35−5{1+5βn+1+1−5αn+1}
Commented by maxmathsup by imad last updated on 01/Sep/18
3) we have f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n ⇒ f(x) = Σ_(n=0) ^∞ {−1 + (1/(3(√5)−5))( ((1+(√5))/β^(n+1) ) +((1−(√5))/α^(n+1) ))}x^n .
3)wehavef(x)=∑n=0∞f(n)(0)n!xn⇒f(x)=∑n=0∞{−1+135−5(1+5βn+1+1−5αn+1)}xn.

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