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Question Number 42689 by prof Abdo imad last updated on 31/Aug/18
let f(x) = (x/(x^3 −2x +1)) 1) find D_f 2) find f^((n)) (x) then f^((n)) (0) 3) developp f at integr serie.
$${let}\:{f}\left({x}\right)\:=\:\frac{{x}}{{x}^{\mathrm{3}} −\mathrm{2}{x}\:\:+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{f}^{\left({n}\right)} \left({x}\right)\:\:{then}\:\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by maxmathsup by imad last updated on 01/Sep/18
1) x∈ D_f ⇔ x^3 −2x +1 ≠0 but x^3 −2x +1 =x^3 −1 −2x +2 =(x−1)(x^2 +x+1)−2(x−1) =(x−1)(x^2 +x+1 −2) =(x−1)(x^2 +x−1) roots of x^2 +x−1? Δ=1−4(−1)=5 ⇒α =((−1+(√5))/2) and β =((−1−(√5))/2) ⇒ D_f =R−{1,((−1+(√5))/2) ,((−1−(√5))/2)}
$$\left.\mathrm{1}\right)\:{x}\in\:{D}_{{f}} \:\Leftrightarrow\:{x}^{\mathrm{3}} −\mathrm{2}{x}\:+\mathrm{1}\:\neq\mathrm{0}\:\:{but} \\ $$$${x}^{\mathrm{3}} −\mathrm{2}{x}\:+\mathrm{1}\:={x}^{\mathrm{3}} −\mathrm{1}\:−\mathrm{2}{x}\:+\mathrm{2}\:=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)−\mathrm{2}\left({x}−\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\:−\mathrm{2}\right)\:=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{x}−\mathrm{1}\right)\:\:{roots}\:{of}\:{x}^{\mathrm{2}} \:+{x}−\mathrm{1}? \\ $$$$\Delta=\mathrm{1}−\mathrm{4}\left(−\mathrm{1}\right)=\mathrm{5}\:\Rightarrow\alpha\:=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:{and}\:\beta\:=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\Rightarrow \\ $$$${D}_{{f}} ={R}−\left\{\mathrm{1},\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:,\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 01/Sep/18
2) we have f(x)= (x/((x−1)(x−α)(x−β))) =(a/(x−1)) +(b/(x−α)) +(c/(x−β)) a =lim_(x→1) (x−1)f(x) = (1/((1−α)(1−β))) b =lim_(x→α) (x−α)f(x) = (α/((α−1)(α−β))) c =lim_(x→β) (x−β)f(x) = (β/((β−1)(β−α))) but α =((−1+(√5))/2) and β=((−1−(√5))/2) ⇒ a = (1/((1−((−1+(√5))/2))(1−((−1−(√5))/2)))) = (4/((3−(√5))(3+(√5)))) =(4/4) =1 b = ((−1+(√5))/(2(1−((−1+(√5))/2))(−(√5)))) =((−1+(√5))/( (√5)(3−(√5)))) =((−1+(√5))/(3(√5)−5)) . c = ((−1−(√5))/(2(1−((−1−(√5))/2))((√5)))) =((−1−(√5))/( (√5)(3−(√5)))) =((−1−(√5))/(3(√5)−5)) ⇒ f(x)= (1/(x−1)) +(1/(3(√5)−5)){ ((−1+(√5))/(x−α)) −((1+(√5))/(x−β))} ⇒ f^((n)) (x) = (((−1)^n n!)/((x−1)^(n+1) )) + ((−1+(√5))/(3(√5)−5)) (((−1)^n n!)/((x−α)^(n+1) )) −((1+(√5))/(3(√5)−5)) (((−1)^n n!)/((x−β)^(n+1) )) ⇒ f^((n)) (0) =−n! +((−1+(√5))/(3(√5)−5)) (((−1)^n n!)/((−α)^(n+1) )) −((1+(√5))/(3(√5)−5)) (((−1)^n n!)/((−β)^(n+1) )) =−n! −((−1+(√5))/(3(√5)−5)) ((n!)/α^(n+1) ) +((1+(√5))/(3(√5)−5)) ((n!)/β^(n+1) ) ⇒ f^((n)) (0) =−n! +((n!)/(3(√5)−5)){ ((1+(√5))/β^(n+1) ) +((1−(√5))/α^(n+1) )}
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}\right)=\:\frac{{x}}{\left({x}−\mathrm{1}\right)\left({x}−\alpha\right)\left({x}−\beta\right)}\:=\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{{x}−\alpha}\:+\frac{{c}}{{x}−\beta} \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right){f}\left({x}\right)\:=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−\alpha\right)\left(\mathrm{1}−\beta\right)} \\ $$$${b}\:={lim}_{{x}\rightarrow\alpha} \left({x}−\alpha\right){f}\left({x}\right)\:=\:\frac{\alpha}{\left(\alpha−\mathrm{1}\right)\left(\alpha−\beta\right)} \\ $$$${c}\:={lim}_{{x}\rightarrow\beta} \left({x}−\beta\right){f}\left({x}\right)\:=\:\frac{\beta}{\left(\beta−\mathrm{1}\right)\left(\beta−\alpha\right)}\:\:{but}\:\alpha\:=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:{and}\:\beta=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow \\ $$$${a}\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\mathrm{1}−\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{4}}{\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)}\:=\frac{\mathrm{4}}{\mathrm{4}}\:=\mathrm{1} \\ $$$${b}\:=\:\:\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}\left(\mathrm{1}−\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(−\sqrt{\mathrm{5}}\right)}\:\:=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}\:=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\:. \\ $$$${c}\:=\:\:\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}\left(\mathrm{1}−\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\sqrt{\mathrm{5}}\right)}\:=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}\:=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:\:+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\left\{\:\:\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{{x}−\alpha}\:−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{{x}−\beta}\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\:+\:\:\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\alpha\right)^{{n}+\mathrm{1}} }\:−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\beta\right)^{{n}+\mathrm{1}} }\:\:\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=−{n}!\:\:\:\:+\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left(−\alpha\right)^{{n}+\mathrm{1}} }\:−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left(−\beta\right)^{{n}+\mathrm{1}} } \\ $$$$=−{n}!\:\:−\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\:\frac{{n}!}{\alpha^{{n}+\mathrm{1}} }\:+\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\:\frac{{n}!}{\beta^{{n}+\mathrm{1}} }\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=−{n}!\:\:+\frac{{n}!}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\left\{\:\:\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\beta^{{n}+\mathrm{1}} }\:+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\alpha^{{n}+\mathrm{1}} }\right\} \\ $$
Commented by maxmathsup by imad last updated on 01/Sep/18
3) we have f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n ⇒ f(x) = Σ_(n=0) ^∞ {−1 + (1/(3(√5)−5))( ((1+(√5))/β^(n+1) ) +((1−(√5))/α^(n+1) ))}x^n .
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \left\{−\mathrm{1}\:\:+\:\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}}\left(\:\:\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\beta^{{n}+\mathrm{1}} }\:+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\alpha^{{n}+\mathrm{1}} }\right)\right\}{x}^{{n}} \:. \\ $$

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