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Question Number 88032 by mathmax by abdo last updated on 07/Apr/20
decompose inside R(x) the fraction 1) F(x) =(1/(x^3 (x−2)^3 )) 2) F(x) =(1/((x+1)^4 (x−3)^4 ))
decomposeinsideR(x)thefraction1)F(x)=1x3(x−2)32)F(x)=1(x+1)4(x−3)4
Commented by jagoll last updated on 08/Apr/20
what is R(x)? remainder?
whatisR(x)?remainder?
Commented by abdomathmax last updated on 10/Apr/20
space of rational fractions
spaceofrationalfractions
Commented by abdomathmax last updated on 10/Apr/20
1)F(x) =(1/8)(((x−(x−2))^3 )/(x^3 (x−2)^3 )) ⇒ 8F(x)?=((x^3 −3x^2 (x−2) +3x(x−2)^2 −(x−2)^3 )/(x^3 (x−2)^3 )) =((x^3 +3x(x−2)(−x+x−2) −(x−2)^3 )/(x^3 (x−2)^3 )) =((x^3 −6x(x−2)−(x−2)^3 )/(x^3 (x−2)^3 )) =(1/((x−2)^3 )) −(6/(x^2 (x−2)^2 ))−(1/x^3 ) =(1/((x−2)^3 ))−(1/x^3 ) −(3/2){(((x−(x−2))^2 )/(x^2 (x−2)^2 ))} =(1/((x−2)^3 ))−(1/x^3 )−(3/2){ ((x^2 −2x(x−2) +(x−2)^2 )/(x^2 (x−2)^2 ))} =(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 )) +(6/(x(x−2))) +(3/(2x^2 )) =(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 )) −3((1/x)−(1/(x−2))) +(3/(2x^2 )) =−(3/x) +(3/(x−2)) +(3/(2x^2 ))−(3/(2(x−2)^2 ))−(1/x^3 ) +(1/((x−2)^3 )) =8F(x)
1)F(x)=18(x−(x−2))3x3(x−2)3⇒8F(x)?=x3−3x2(x−2)+3x(x−2)2−(x−2)3x3(x−2)3=x3+3x(x−2)(−x+x−2)−(x−2)3x3(x−2)3=x3−6x(x−2)−(x−2)3x3(x−2)3=1(x−2)3−6x2(x−2)2−1x3=1(x−2)3−1x3−32{(x−(x−2))2x2(x−2)2}=1(x−2)3−1x3−32{x2−2x(x−2)+(x−2)2x2(x−2)2}=1(x−2)3−1x3−32(x−2)2+6x(x−2)+32x2=1(x−2)3−1x3−32(x−2)2−3(1x−1x−2)+32x2=−3x+3x−2+32x2−32(x−2)2−1x3+1(x−2)3=8F(x)
Commented by mathmax by abdo last updated on 10/Apr/20
error of calculus 8F(x)=(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 )) +(3/(x(x−2)))−(3/(2x^2 )) =(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 )) −(3/2)((1/x)−(1/(x−2)))−(3/(2x^2 )) F(x)=(1/8){(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 ))−(3/(2x)) +(3/(2(x−2)))−(3/(2x^2 ))}
errorofcalculus8F(x)=1(x−2)3−1x3−32(x−2)2+3x(x−2)−32x2=1(x−2)3−1x3−32(x−2)2−32(1x−1x−2)−32x2F(x)=18{1(x−2)3−1x3−32(x−2)2−32x+32(x−2)−32x2}
Commented by mathmax by abdo last updated on 10/Apr/20
2)F(x)=(1/4^4 )(({(x+1)−(x−3)}^4 )/((x+1)^4 (x−3)^4 )) ⇒4^4 F(x)=(({(x+1)^2 −2(x+1)(x−3)+(x−3)^2 }^2 )/((x+1)^4 (x−3)^4 )) we use the identity (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab +bc +ca) ⇒ 4^4 F(x)=(((x+1)^4 +4(x+1)^2 (x−3)^2 +(x−3)^4 +2{ −2(x+1)^3 (x−3)+(x+1)^2 (x−3)^2 −2(x+1)(x−3)^3 })/((x+1)^4 (x−3)^4 )) =(1/((x−3)^4 )) +(4/((x+1)^2 (x−3)^2 )) +(1/((x+1)^4 )) −(4/((x+1)(x−3)^3 )) +(2/((x+1)^2 (x−3)^2 ))−(4/((x+1)^3 (x−3))) also we have (4/((x+1)^2 (x−3)^2 )) =(1/4){((((x+1)−(x−3))^2 )/((x+1)^2 (x−3)^2 ))} =(1/4){(((x+1)^2 −2(x+1)(x−3) +(x−3)^2 )/((x+1)^2 (x−3)^2 ))} =(1/4){ (1/((x−3)^2 )) −(2/((x+1)(x−3))) +(1/((x+1)^2 ))} =(1/(4(x−3)^2 ))+(1/8)((1/(x+1))−(1/(x−3)))+(1/(4(x+1)^2 )) =(1/(4(x−3)^2 )) +(1/(8(x+1)))−(1/(8(x−3)))+(1/(4(x+1)^2 )) −(4/((x+1)(x−3)^3 ))−(4/((x+1)^3 (x−3))) =−(4/((x+1)(x−3)))((1/((x−3)^2 ))+(1/((x+1)^2 ))) =((1/(x+1))−(1/(x−3)))((1/((x−3)^2 ))+(1/((x+1)^2 ))) =(1/((x+1)(x−3)^2 )) +(1/((x+1)^3 ))−(1/((x−3)^3 ))−(1/((x−3)(x+1)^2 )) =(1/((x+1)(x−3))){(1/(x−3))−(1/(x+1))}+(1/((x+1)^3 ))−(1/((x−3)^3 )) =−(1/4)((1/(x+1))−(1/(x−3)))((1/(x−3))−(1/(x+1)))+(1/((x+1)^3 ))−(1/((x−3)^3 )) =−(1/4)((1/((x+1)(x−3)))−(1/((x+1)^2 ))−(1/((x−3)^2 ))+(1/((x−3)(x+1))))+(1/((x+1)^3 ))−(1/((x−3)^3 )) ....
2)F(x)=144{(x+1)−(x−3)}4(x+1)4(x−3)4⇒44F(x)={(x+1)2−2(x+1)(x−3)+(x−3)2}2(x+1)4(x−3)4weusetheidentity(a+b+c)2=a2+b2+c2+2(ab+bc+ca)⇒44F(x)=(x+1)4+4(x+1)2(x−3)2+(x−3)4+2{−2(x+1)3(x−3)+(x+1)2(x−3)2−2(x+1)(x−3)3}(x+1)4(x−3)4=1(x−3)4+4(x+1)2(x−3)2+1(x+1)4−4(x+1)(x−3)3+2(x+1)2(x−3)2−4(x+1)3(x−3)alsowehave4(x+1)2(x−3)2=14{((x+1)−(x−3))2(x+1)2(x−3)2}=14{(x+1)2−2(x+1)(x−3)+(x−3)2(x+1)2(x−3)2}=14{1(x−3)2−2(x+1)(x−3)+1(x+1)2}=14(x−3)2+18(1x+1−1x−3)+14(x+1)2=14(x−3)2+18(x+1)−18(x−3)+14(x+1)2−4(x+1)(x−3)3−4(x+1)3(x−3)=−4(x+1)(x−3)(1(x−3)2+1(x+1)2)=(1x+1−1x−3)(1(x−3)2+1(x+1)2)=1(x+1)(x−3)2+1(x+1)3−1(x−3)3−1(x−3)(x+1)2=1(x+1)(x−3){1x−3−1x+1}+1(x+1)3−1(x−3)3=−14(1x+1−1x−3)(1x−3−1x+1)+1(x+1)3−1(x−3)3=−14(1(x+1)(x−3)−1(x+1)2−1(x−3)2+1(x−3)(x+1))+1(x+1)3−1(x−3)3….
Commented by abdomathmax last updated on 10/Apr/20
1) another way we detemine D_2 f(0) with f(x)=(1/((x−2)^3 )) ⇒f(x)=f(0) +(x/(1!))f^((1)) (0)+(x^2 /(2!))f^((2)) (0) +x^3 ξ(x) f(0)=−(1/8) f(x)=(x−2)^(−3) ⇒f^′ (x)=−3(x−2)^(−4) ⇒ f^′ (0)=−3(−2)^(−4) =((−3)/2^4 ) =−(3/(16)) f^((2)) (0) =12(x−2)^(−5) ⇒f^((2)) (0) =((12)/((−2)^5 )) =−((12)/2^5 ) =−((12)/(32)) =−(3/8) decomposition of F is F(x)=Σ_(i=1) ^3 (a_i /x^i ) +Σ_(i=1) ^3 (b_i /((x−2)^i )) we have F(x)=(1/x^3 )(−(1/8)−(3/(16))x−(3/(16))x^2 +x^3 ξ(x)) =−(1/(8x^3 ))−(3/(16x^2 ))−(3/(16x)) +ξ(x) ⇒a_1 =−(3/(16)) a_2 =−(3/(16)) and a_3 =−(1/8) let find the b_i we do the changement x−2 =t ⇒ F(x)=(1/(t^3 (t+2)^3 )) and find D_2 (0) for g(t)=(t+2)^(−3) g(t)=g(0)+t g^′ (0) +(t^2 /2)g^((2)) (0) +t^3 δ(x) g(0)=(1/8) , g^′ (t)=−3(t+2)^(−4) ⇒g^′ (0)=((−3)/(16)) g^((2)) (0) =12(t+2)^(−5) ⇒g^((2)) (0)=((12)/2^5 ) =((12)/(32)) =(3/8) F(x)=(1/t^3 ){(1/8)−(3/(16))t +(3/(16))t^2 +t^3 δ(t)} =(1/(8t^3 )) −(3/(16t^2 )) +(3/(16t)) +δ(t) =(1/(8(x−2)^3 ))−(3/(16(x−2)^2 ))+(3/(16(x−2))) +δ(t) ⇒ b_1 =(3/(16)) , b_2 =−(3/(26)) and b_3 =(1/8) tbe coefficients are found
1)anotherwaywedetemineD2f(0)withf(x)=1(x−2)3⇒f(x)=f(0)+x1!f(1)(0)+x22!f(2)(0)+x3ξ(x)f(0)=−18f(x)=(x−2)−3⇒f′(x)=−3(x−2)−4⇒f′(0)=−3(−2)−4=−324=−316f(2)(0)=12(x−2)−5⇒f(2)(0)=12(−2)5=−1225=−1232=−38decompositionofFisF(x)=∑i=13aixi+∑i=13bi(x−2)iwehaveF(x)=1x3(−18−316x−316x2+x3ξ(x))=−18x3−316x2−316x+ξ(x)⇒a1=−316a2=−316anda3=−18letfindthebiwedothechangementx−2=t⇒F(x)=1t3(t+2)3andfindD2(0)forg(t)=(t+2)−3g(t)=g(0)+tg′(0)+t22g(2)(0)+t3δ(x)g(0)=18,g′(t)=−3(t+2)−4⇒g′(0)=−316g(2)(0)=12(t+2)−5⇒g(2)(0)=1225=1232=38F(x)=1t3{18−316t+316t2+t3δ(t)}=18t3−316t2+316t+δ(t)=18(x−2)3−316(x−2)2+316(x−2)+δ(t)⇒b1=316,b2=−326andb3=18tbecoefficientsarefound
Commented by abdomathmax last updated on 10/Apr/20
b_2 =−(3/(16))
b2=−316

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