Menu Close

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-

Tinku Tara 0 Comments
Pin




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 ))
$${decompose}\:{inside}\:{R}\left({x}\right)\:{the}\:{fraction} \\ $$$$\left.\mathrm{1}\right)\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} \left({x}−\mathrm{3}\right)^{\mathrm{4}} } \\ $$
Commented by jagoll last updated on 08/Apr/20
what is R(x)? remainder?
$$\mathrm{what}\:\mathrm{is}\:\mathrm{R}\left(\mathrm{x}\right)?\:\mathrm{remainder}? \\ $$
Commented by abdomathmax last updated on 10/Apr/20
space of rational fractions
$${space}\:{of}\:{rational}\:{fractions} \\ $$
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)
$$\left.\mathrm{1}\right){F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}\frac{\left({x}−\left({x}−\mathrm{2}\right)\right)^{\mathrm{3}} }{{x}^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\mathrm{8}{F}\left({x}\right)?=\frac{{x}^{\mathrm{3}} \:−\mathrm{3}{x}^{\mathrm{2}} \left({x}−\mathrm{2}\right)\:+\mathrm{3}{x}\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:−\left({x}−\mathrm{2}\right)^{\mathrm{3}} }{{x}^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{3}} } \\ $$$$=\frac{{x}^{\mathrm{3}} \:+\mathrm{3}{x}\left({x}−\mathrm{2}\right)\left(−{x}+{x}−\mathrm{2}\right)\:−\left({x}−\mathrm{2}\right)^{\mathrm{3}} }{{x}^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{3}} } \\ $$$$=\frac{{x}^{\mathrm{3}} \:−\mathrm{6}{x}\left({x}−\mathrm{2}\right)−\left({x}−\mathrm{2}\right)^{\mathrm{3}} }{{x}^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }\:−\frac{\mathrm{6}}{{x}^{\mathrm{2}} \left({x}−\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:−\frac{\mathrm{3}}{\mathrm{2}}\left\{\frac{\left({x}−\left({x}−\mathrm{2}\right)\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}−\mathrm{2}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{3}}{\mathrm{2}}\left\{\:\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}\left({x}−\mathrm{2}\right)\:+\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}−\mathrm{2}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:+\frac{\mathrm{6}}{{x}\left({x}−\mathrm{2}\right)}\:+\frac{\mathrm{3}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:−\mathrm{3}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}−\mathrm{2}}\right)\:+\frac{\mathrm{3}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{3}}{{x}}\:+\frac{\mathrm{3}}{{x}−\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{2}{x}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }\:=\mathrm{8}{F}\left({x}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
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 ))}
$${error}\:{of}\:{calculus} \\ $$$$\mathrm{8}{F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:+\frac{\mathrm{3}}{{x}\left({x}−\mathrm{2}\right)}−\frac{\mathrm{3}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}−\mathrm{2}}\right)−\frac{\mathrm{3}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left\{\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{2}{x}}\:+\frac{\mathrm{3}}{\mathrm{2}\left({x}−\mathrm{2}\right)}−\frac{\mathrm{3}}{\mathrm{2}{x}^{\mathrm{2}} }\right\} \\ $$
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 )) ....
$$\left.\mathrm{2}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{4}} }\frac{\left\{\left({x}+\mathrm{1}\right)−\left({x}−\mathrm{3}\right)\right\}^{\mathrm{4}} }{\left({x}+\mathrm{1}\right)^{\mathrm{4}} \left({x}−\mathrm{3}\right)^{\mathrm{4}} } \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{4}} \:{F}\left({x}\right)=\frac{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)+\left({x}−\mathrm{3}\right)^{\mathrm{2}} \right\}^{\mathrm{2}} }{\left({x}+\mathrm{1}\right)^{\mathrm{4}} \left({x}−\mathrm{3}\right)^{\mathrm{4}} } \\ $$$${we}\:{use}\:{the}\:{identity}\:\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:+{c}^{\mathrm{2}} \:+\mathrm{2}\left({ab}\:+{bc}\:+{ca}\right)\:\Rightarrow \\ $$$$\mathrm{4}^{\mathrm{4}} \:{F}\left({x}\right)=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{4}} \:+\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} \:+\left({x}−\mathrm{3}\right)^{\mathrm{4}} \:+\mathrm{2}\left\{\:−\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}−\mathrm{3}\right)+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)^{\mathrm{3}} \right\}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} \left({x}−\mathrm{3}\right)^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{4}} }\:+\frac{\mathrm{4}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }\:−\frac{\mathrm{4}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)^{\mathrm{3}} }\:+\frac{\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} }−\frac{\mathrm{4}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}−\mathrm{3}\right)} \\ $$$${also}\:{we}\:{have}\:\frac{\mathrm{4}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\left\{\frac{\left(\left({x}+\mathrm{1}\right)−\left({x}−\mathrm{3}\right)\right)^{\mathrm{2}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)\:+\left({x}−\mathrm{3}\right)^{\mathrm{2}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:−\frac{\mathrm{2}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)}\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{{x}−\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{8}\left({x}−\mathrm{3}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{4}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)^{\mathrm{3}} }−\frac{\mathrm{4}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}−\mathrm{3}\right)}\:=−\frac{\mathrm{4}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)}\left(\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{{x}−\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)}\left\{\frac{\mathrm{1}}{{x}−\mathrm{3}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right\}+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{3}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{{x}−\mathrm{3}}\right)\left(\frac{\mathrm{1}}{{x}−\mathrm{3}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{3}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)}−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{1}\right)}\right)+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{3}} } \\ $$$$…. \\ $$
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
$$\left.\mathrm{1}\right)\:{another}\:{way}\:\:{we}\:{detemine}\:{D}_{\mathrm{2}} {f}\left(\mathrm{0}\right)\:{with} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} }\:\Rightarrow{f}\left({x}\right)={f}\left(\mathrm{0}\right)\:+\frac{{x}}{\mathrm{1}!}{f}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right) \\ $$$$+{x}^{\mathrm{3}} \xi\left({x}\right) \\ $$$${f}\left(\mathrm{0}\right)=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{2}\right)^{−\mathrm{3}} \:\Rightarrow{f}^{'} \left({x}\right)=−\mathrm{3}\left({x}−\mathrm{2}\right)^{−\mathrm{4}} \:\Rightarrow \\ $$$${f}^{'} \left(\mathrm{0}\right)=−\mathrm{3}\left(−\mathrm{2}\right)^{−\mathrm{4}} \:=\frac{−\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }\:=−\frac{\mathrm{3}}{\mathrm{16}} \\ $$$${f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\mathrm{12}\left({x}−\mathrm{2}\right)^{−\mathrm{5}} \:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\frac{\mathrm{12}}{\left(−\mathrm{2}\right)^{\mathrm{5}} }\:=−\frac{\mathrm{12}}{\mathrm{2}^{\mathrm{5}} } \\ $$$$=−\frac{\mathrm{12}}{\mathrm{32}}\:=−\frac{\mathrm{3}}{\mathrm{8}}\:\:{decomposition}\:{of}\:{F}\:{is} \\ $$$${F}\left({x}\right)=\sum_{{i}=\mathrm{1}} ^{\mathrm{3}} \:\frac{{a}_{{i}} }{{x}^{{i}} }\:+\sum_{{i}=\mathrm{1}} ^{\mathrm{3}} \:\frac{{b}_{{i}} }{\left({x}−\mathrm{2}\right)^{{i}} }\:\:{we}\:{have} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\left(−\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{3}}{\mathrm{16}}{x}−\frac{\mathrm{3}}{\mathrm{16}}{x}^{\mathrm{2}} \:+{x}^{\mathrm{3}} \xi\left({x}\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{3}} }−\frac{\mathrm{3}}{\mathrm{16}{x}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{16}{x}}\:+\xi\left({x}\right)\:\Rightarrow{a}_{\mathrm{1}} =−\frac{\mathrm{3}}{\mathrm{16}} \\ $$$${a}_{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{16}}\:\:{and}\:{a}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{8}}\:\:{let}\:{find}\:{the}\:{b}_{{i}} \:{we}\:{do}\:{the} \\ $$$${changement}\:{x}−\mathrm{2}\:={t}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{3}} \left({t}+\mathrm{2}\right)^{\mathrm{3}} }\:\:{and}\:{find}\:{D}_{\mathrm{2}} \left(\mathrm{0}\right)\:{for}\:{g}\left({t}\right)=\left({t}+\mathrm{2}\right)^{−\mathrm{3}} \\ $$$${g}\left({t}\right)={g}\left(\mathrm{0}\right)+{t}\:{g}^{'} \left(\mathrm{0}\right)\:+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}{g}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:+{t}^{\mathrm{3}} \delta\left({x}\right) \\ $$$${g}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{8}}\:\:,\:\:\:{g}^{'} \left({t}\right)=−\mathrm{3}\left({t}+\mathrm{2}\right)^{−\mathrm{4}} \:\Rightarrow{g}^{'} \left(\mathrm{0}\right)=\frac{−\mathrm{3}}{\mathrm{16}} \\ $$$${g}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\mathrm{12}\left({t}+\mathrm{2}\right)^{−\mathrm{5}} \:\Rightarrow{g}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\frac{\mathrm{12}}{\mathrm{2}^{\mathrm{5}} }\:=\frac{\mathrm{12}}{\mathrm{32}}\:=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\left\{\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{3}}{\mathrm{16}}{t}\:+\frac{\mathrm{3}}{\mathrm{16}}{t}^{\mathrm{2}} \:+{t}^{\mathrm{3}} \delta\left({t}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{t}^{\mathrm{3}} }\:−\frac{\mathrm{3}}{\mathrm{16}{t}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{16}{t}}\:+\delta\left({t}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}\left({x}−\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{3}}{\mathrm{16}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{16}\left({x}−\mathrm{2}\right)}\:+\delta\left({t}\right)\:\Rightarrow \\ $$$${b}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{16}}\:,\:{b}_{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{26}}\:\:{and}\:{b}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${tbe}\:{coefficients}\:{are}\:{found} \\ $$$$ \\ $$
Commented by abdomathmax last updated on 10/Apr/20
b_2 =−(3/(16))
$${b}_{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{16}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *