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decompose-F-x-1-1-x-2-1-x-2-inside-R-x-

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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).
decomposeF(x)=1(1−x)2(1−x2)insideR(x).
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(x)=1(1−x)2(1−x)(1+x)=1(1−x)3(1+x)=−1(x−1)3(x+1)=ax−1+b(x−1)2+c(x−1)3+dx+1c=limx→1(x−1)3F(x)=−12d=limx→−1(x+1)F(x)=18⇒F(x)=ax−1+b(x−1)2−12(x−1)3+18(x+1)F(0)=1=−a+b+12+18=−a+b+58⇒−a+b=1−58=38F(2)=−13=a+b−12+124=a+b−1124⇒a+b=−13+1124=324=18wegetthesystem−a+b=38anda+b=18⇒2b=12⇒b=14a=18−b=18−14=−18andF(x)=−18(x−1)+14(x−1)2−12(x−1)3+18(x+1).

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