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Question Number 35988 by abdo mathsup 649 cc last updated on 26/May/18
let f(x) = ((x+2)/(x^3 −4x +3)) 1) calculate f^((n)) (x) 2) developp f at integr serie.
letf(x)=x+2x3−4x+31)calculatef(n)(x)2)developpfatintegrserie.
Commented by abdo mathsup 649 cc last updated on 27/May/18
1) we have x^3 −4x +3 =x^3 −x −3x +3 =x(x^2 −1) −3(x−1) = (x−1)( x^2 +x −3) ⇒ f(x) = ((x+2)/((x−1)(x^2 +x−3))) = (a/(x−1)) +((bx +c)/(x^2 +x−3)) a=lim_(x→1) (x−1)f(x) = (3/(−1)) =−3 lim_(x→+∞) x f(x)=0 = a +b ⇒b=−a = 3 ⇒ f(x) = ((−3)/(x−1)) + ((3x+c)/(x^(2 ) +x −3)) f(0) = (2/3) = 3 −(c/3) ⇒2= 9 −c ⇒ c =7 ⇒ f(x) = ((−3)/(x−1)) +((3x+7)/(x^2 +x−3)) let find the roots of x^2 +x−3 , Δ =1−4(−3) =13 x_(1 ) = ((−1+(√(13)))/2) , x_2 = ((−1−(√(13)))/2) so ((3x+7)/(x^2 +x−3)) = ((3x+7)/((x−x_1 )(x−x_2 ))) =((3(x−x_1 ) +3x_1 +7)/((x−x_1 )(x−x_2 ))) = (3/(x−x_2 )) + ((3x_(1 ) +7)/((x−x_1 )(x−x_2 ))) (3/(x−x_2 )) + ((3x_1 +7)/(x_1 −x_2 )) { (1/(x−x_1 )) −(1/(x−x_2 ))} ⇒ f(x) = −(3/(x−1)) +(3/(x−x_2 )) + (λ/(x−x_1 )) −(λ/(x−x_2 )) with λ = ((3x_(1 ) +7)/(x_1 −x_2 )) ⇒ f^((n)) (x) = −3 (((−1)^n n!)/((x−1)^(n+1) )) + (3−λ)(((−1)^n n!)/((x−x_2 )^(n+1) )) +λ (((−1)^n n!)/((x−x_1 )^(n+1) ))
1)wehavex3−4x+3=x3−x−3x+3=x(x2−1)−3(x−1)=(x−1)(x2+x−3)⇒f(x)=x+2(x−1)(x2+x−3)=ax−1+bx+cx2+x−3a=limx→1(x−1)f(x)=3−1=−3limx→+∞xf(x)=0=a+b⇒b=−a=3⇒f(x)=−3x−1+3x+cx2+x−3f(0)=23=3−c3⇒2=9−c⇒c=7⇒f(x)=−3x−1+3x+7x2+x−3letfindtherootsofx2+x−3,Δ=1−4(−3)=13x1=−1+132,x2=−1−132so3x+7x2+x−3=3x+7(x−x1)(x−x2)=3(x−x1)+3x1+7(x−x1)(x−x2)=3x−x2+3x1+7(x−x1)(x−x2)3x−x2+3x1+7x1−x2{1x−x1−1x−x2}⇒f(x)=−3x−1+3x−x2+λx−x1−λx−x2withλ=3x1+7x1−x2⇒f(n)(x)=−3(−1)nn!(x−1)n+1+(3−λ)(−1)nn!(x−x2)n+1+λ(−1)nn!(x−x1)n+1
Commented by abdo mathsup 649 cc last updated on 27/May/18
2) we have f^((n)) (0) = 3n! +(3−λ) ((n!)/(−x_2 ^(n+1) )) +λ ((n!)/(−x_1 ^(n+1) )) f^((n)) (0) = 3(n!) +(λ −3) ((n!)/x_2 ^(n+1) ) −λ ((n!)/x_1 ^(n+1) ) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n = Σ_(n=0) ^∞ (3 +((λ−3)/x_2 ^(n+1) ) −(λ/x_1 ^(n+1) )) x^n .
2)wehavef(n)(0)=3n!+(3−λ)n!−x2n+1+λn!−x1n+1f(n)(0)=3(n!)+(λ−3)n!x2n+1−λn!x1n+1f(x)=∑n=0∞f(n)(0)n!xn=∑n=0∞(3+λ−3x2n+1−λx1n+1)xn.
Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18
Dr=x^3 −4x+4−1 =x^3 −1−4(x−1) =(x−1)(x^2 +x+1)−4(x−1) (x−1)(x^2 +x+1−4) =(x−1)(x^2 +x−3) ((x+2)/((x−1)(x^2 +x−3)))=(A/(x−1))+((Bx+C)/((x^2 +x−3))) x+2=A(x^2 +x−3)+(x−1)(Bx+C) put x=1 3=−A so A=−3 x+2=−3x^2 −3x+9+Bx^2 +Cx−Bx−C x+2=x^2 (−3+B)+x(−3−B+C)+9−C −3+B=0 B=3 −3−B+C=1 −3−3+C=1 C=7 f(x)=((−3)/(x−1))+((3x+7)/(x^2 +x−3)) contf f(x)=u+v (u/(−3))=(x−1)^(−1) (u_1 /(−3))=(−1)(x−1)^(−2) (u_2 /(−3))=(−1)(−2)(x−1)^(−3) (u_3 /(−3))=(−1)(−2)(−3)(x−1)^(−4) ..... ..... (u_n /(−3))=(−1)(−2)(−3)...(−n)(x−1)^(−n−1) (u_n /(−3))=(((−)^n ×n!)/((x−1)^(n+1) )) u_n =(−3)×(((−1)^n ×n!)/((x−1)^(n+1) )) pls check contd
Dr=x3−4x+4−1=x3−1−4(x−1)=(x−1)(x2+x+1)−4(x−1)(x−1)(x2+x+1−4)=(x−1)(x2+x−3)x+2(x−1)(x2+x−3)=Ax−1+Bx+C(x2+x−3)x+2=A(x2+x−3)+(x−1)(Bx+C)putx=13=−AsoA=−3x+2=−3x2−3x+9+Bx2+Cx−Bx−Cx+2=x2(−3+B)+x(−3−B+C)+9−C−3+B=0B=3−3−B+C=1−3−3+C=1C=7f(x)=−3x−1+3x+7x2+x−3contff(x)=u+vu−3=(x−1)−1u1−3=(−1)(x−1)−2u2−3=(−1)(−2)(x−1)−3u3−3=(−1)(−2)(−3)(x−1)−4…..…..un−3=(−1)(−2)(−3)…(−n)(x−1)−n−1un−3=(−)n×n!(x−1)n+1un=(−3)×(−1)n×n!(x−1)n+1plscheckcontd
Commented by behi83417@gmail.com last updated on 26/May/18
dear mr tanmay and mr Rasheed! a comment have posted to Q.35844 please check it.thanks.
dearmrtanmayandmrRasheed!acommenthavepostedtoQ.35844pleasecheckit.thanks.

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