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Question Number 67672 by Abdo msup. last updated on 30/Aug/19
decompose the folowing fraction at R(x) 1)F(x)=(x^3 /(1−x^6 )) 2) G(x) =((x^2 +1)/(x^3 (x^2 +x+1)^2 ))
$${decompose}\:{the}\:{folowing}\:\:{fraction}\:{at}\:{R}\left({x}\right) \\ $$$$\left.\mathrm{1}\right){F}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{6}} } \\ $$$$\left.\mathrm{2}\right)\:{G}\left({x}\right)\:=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by Abdo msup. last updated on 30/Aug/19
1)F(x) =−(x^3 /((x^3 )^2 −1)) =−(x^3 /((x^3 −1)(x^3 +1))) =((−x^3 )/((x−1)(x^2 +x+1)(x+1)(x^2 −x+1))) =(a/(x−1)) +(b/(x+1)) +((cx +d)/(x^2 +x+1)) +((ex +f)/(x^2 −x +1)) a =lim_(x→1) (x−1)F(x) =−(1/(3.2.1)) =−(1/6) b =lim_(x→−1) (x+1)F(x)=(1/((−2).3))=−(1/6) ⇒ F(x)=((−1)/(6(x−1)))−(1/(6(x+1))) +((cx+d)/(x^2 +x+1)) +((ex +f)/(x^2 −x+1)) F(−x) =−F(x) ⇒ (1/(6(x+1))) +(1/(6(x−1))) +((−cx+d)/(x^2 −x +1)) +((−ex +f)/(x^2 +x+1)) =(1/(6(x−1))) +(1/(6(x+1))) +((−cx −d)/(x^2 +x+1)) +((−ex−f)/(x^2 −x +1)) ⇒ ⇒ f=−d...
$$\left.\mathrm{1}\right){F}\left({x}\right)\:=−\frac{{x}^{\mathrm{3}} }{\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{1}}\:=−\frac{{x}^{\mathrm{3}} }{\left({x}^{\mathrm{3}} −\mathrm{1}\right)\left({x}^{\mathrm{3}} \:+\mathrm{1}\right)} \\ $$$$=\frac{−{x}^{\mathrm{3}} }{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)} \\ $$$$=\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{cx}\:+{d}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:+\frac{{ex}\:+{f}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right){F}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}\:=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${b}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\left(−\mathrm{2}\right).\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{−\mathrm{1}}{\mathrm{6}\left({x}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({x}+\mathrm{1}\right)}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:+\frac{{ex}\:+{f}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$${F}\left(−{x}\right)\:=−{F}\left({x}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{6}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{6}\left({x}−\mathrm{1}\right)}\:+\frac{−{cx}+{d}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}\:+\frac{−{ex}\:+{f}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{6}\left({x}+\mathrm{1}\right)}\:+\frac{−{cx}\:−{d}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:+\frac{−{ex}−{f}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}\:\Rightarrow \\ $$$$\Rightarrow\:{f}=−{d}… \\ $$$$ \\ $$$$ \\ $$
Commented by Abdo msup. last updated on 30/Aug/19
⇒ F(x) =−(1/(6(x−1)))−(1/(6(x+1))) +((cx +d)/(x^2 +x+1)) +((ex−d)/(x^2 −x+1)) lim_(x→+∞) xF(x) =0 =−(1/3) +c+e ⇒e =(1/3)−c ⇒ F(x) =−(1/(6(x−1)))−(1/(6(x+1))) +((cx+d)/(x^2 +x+1)) +((((1/3)−c)x−d)/(x^2 −x +1)) to detrmine c and d we?can calculate F(2) and F(−2)...be?vontinued...
$$\Rightarrow\:{F}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{6}\left({x}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({x}+\mathrm{1}\right)}\:+\frac{{cx}\:+{d}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:+\frac{{ex}−{d}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)\:=\mathrm{0}\:=−\frac{\mathrm{1}}{\mathrm{3}}\:+{c}+{e}\:\Rightarrow{e}\:=\frac{\mathrm{1}}{\mathrm{3}}−{c}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{6}\left({x}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({x}+\mathrm{1}\right)}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:+\frac{\left(\frac{\mathrm{1}}{\mathrm{3}}−{c}\right){x}−{d}}{{x}^{\mathrm{2}} \:−{x}\:+\mathrm{1}} \\ $$$${to}\:{detrmine}\:{c}\:{and}\:{d}\:\:{we}?{can}\:{calculate}\:{F}\left(\mathrm{2}\right)\:{and} \\ $$$${F}\left(−\mathrm{2}\right)…{be}?{vontinued}… \\ $$

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