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Question Number 32333 by abdo imad last updated on 23/Mar/18
decompose F(x) = (1/((1−x)^2 (1−x^2 ))) inside R(x).
$${decompose}\:{F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:{inside}\:{R}\left({x}\right). \\ $$
Commented by abdo imad last updated on 01/Apr/18
F(x)= (1/((1−x)^2 (1−x)(1+x))) = (1/((1−x)^3 (1+x))) =((−1)/((x−1)^3 (x+1))) = (a/(x−1)) +(b/((x−1)^2 )) +(c/((x−1)^3 )) +(d/(x+1)) c =lim_(x→1) (x−1)^3 F(x)=((−1)/2) d =lim_(x→−1) (x+1)F(x) = (1/8) ⇒ F(x)= (a/(x−1)) +(b/((x−1)^2 )) −(1/(2(x−1)^3 )) + (1/(8(x+1))) F(0) =1 = −a +b +(1/2) +(1/8) =−a+b +(5/8) ⇒−a+b =1−(5/8) =(3/8) F(2)=((−1)/3) = a+b −(1/2) +(1/(24)) =a+b −((11)/(24)) ⇒a+b = −(1/3) +((11)/(24)) = (3/(24)) =(1/8) we get the system −a+b =(3/8) and a+b =(1/8) ⇒ 2b = (1/2) ⇒b=(1/4) a=(1/8) −b =(1/8) −(1/4) = −(1/8) and F(x)= ((−1)/(8(x−1))) +(1/(4(x−1)^2 )) −(1/(2(x−1)^3 ))+(1/(8(x+1))) .
$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \left(\mathrm{1}+{x}\right)}\:=\frac{−\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)} \\ $$$$=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:+\frac{{d}}{{x}+\mathrm{1}} \\ $$$${c}\:={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right)^{\mathrm{3}} {F}\left({x}\right)=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${d}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:\:+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{1}\right)} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\:−{a}\:+{b}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:=−{a}+{b}\:+\frac{\mathrm{5}}{\mathrm{8}} \\ $$$$\Rightarrow−{a}+{b}\:=\mathrm{1}−\frac{\mathrm{5}}{\mathrm{8}}\:=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${F}\left(\mathrm{2}\right)=\frac{−\mathrm{1}}{\mathrm{3}}\:=\:{a}+{b}\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{24}}\:={a}+{b}\:\:−\frac{\mathrm{11}}{\mathrm{24}} \\ $$$$\Rightarrow{a}+{b}\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{11}}{\mathrm{24}}\:=\:\frac{\mathrm{3}}{\mathrm{24}}\:=\frac{\mathrm{1}}{\mathrm{8}}\:{we}\:{get}\:{the}\:{system} \\ $$$$−{a}+{b}\:=\frac{\mathrm{3}}{\mathrm{8}}\:{and}\:\:{a}+{b}\:=\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow\:\mathrm{2}{b}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{8}}\:−{b}\:=\frac{\mathrm{1}}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:=\:−\frac{\mathrm{1}}{\mathrm{8}}\:{and} \\ $$$${F}\left({x}\right)=\:\frac{−\mathrm{1}}{\mathrm{8}\left({x}−\mathrm{1}\right)}\:\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{1}\right)}\:. \\ $$

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