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Question Number 38521 by math khazana by abdo last updated on 26/Jun/18

letf(x) = ((2x+1)/((x−2)(x^2 +x+1))) 1) calculate f^((n)) (x) 2) find f^((n)) (0) 3) developp f at integr serie. (

$${letf}\left({x}\right)\:=\:\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$$$\left(\right. \\ $$

Commented by abdo mathsup 649 cc last updated on 28/Jun/18

1) let drcompose f inside C f(x)= (a/(x−2)) +(b/(x−j)) +(c/(x−j^− )) withj=e^((i2π)/3) f(x)=((2x+1)/((x−2)(x−j)(x−j^− ))) a =lim_(x→2) (x−2)f(x)= (5/7) b=lim_(x→j) (x−j)f(x)= ((2j+1)/((j−2)2i((√3)/2))) =((2j+1)/(i(√3)(j−2))) c=lim_(x→j^− ) (x−j^− )f(x)= ((2j^− +1)/((j^− −2)(−2i((√3)/2)))) =−((2j^− +1)/(i(√3)(j^− −2))) ⇒f(x)= (5/(7(x−2))) +((2j+1)/(i(√3)(j−2)(x−j))) −((2j^(− ) +1)/(i(√3)(j^− −2)(x−j^− ))) ⇒ f^((n)) (x) = (5/7) (((−1)^n n!)/((x−2)^(n+1) )) +((2j+1)/(i(√3)(j−2))) (((−1)^n n!)/((x−j)^(n+1) )) −((2j^− +1)/(i(√3)(j^− −2))) (((−1)^n n!)/((x−j^− )^(n+1) )) but 2)f^((n)) (0) = ((5(−1)^n n!)/(7(−2)^(n+1) )) +((2j+1)/(i(√3)(j−2))) (((−1)^n n!)/((−j)^(n+1) )) −((2j^− +1)/(i(√3)(j^− −2)))(((−1)^n n!)/((−j^− )^(n+1) )) =−(5/7) ((n!)/2^(n+1) ) −((2j+1)/(i(√3)(j−2))) ((n!)/j^(n+1) ) +((2j^− +1)/(i(√3)(j^− −2))) ((n!)/((j^− )^(n+1) )) 3) f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=0) ^∞ { −(5/(7 .2^(n+1) )) −((2j+1)/(i(√3)(j−2)j^(n+1) )) +((2j^− +1)/(i(√3)(j^− −2)(j^− )^(n+1) ))}x^n

$$\left.\mathrm{1}\right)\:{let}\:{drcompose}\:{f}\:{inside}\:{C} \\ $$$${f}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{2}}\:+\frac{{b}}{{x}−{j}}\:+\frac{{c}}{{x}−\overset{−} {{j}}}\:\:{withj}={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}−{j}\right)\left({x}−\overset{−} {{j}}\right)} \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{2}} \left({x}−\mathrm{2}\right){f}\left({x}\right)=\:\frac{\mathrm{5}}{\mathrm{7}} \\ $$$${b}={lim}_{{x}\rightarrow{j}} \:\left({x}−{j}\right){f}\left({x}\right)=\:\frac{\mathrm{2}{j}+\mathrm{1}}{\left({j}−\mathrm{2}\right)\mathrm{2}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\mathrm{2}{j}+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({j}−\mathrm{2}\right)} \\ $$$${c}={lim}_{{x}\rightarrow\overset{−} {{j}}} \left({x}−\overset{−} {{j}}\right){f}\left({x}\right)=\:\frac{\mathrm{2}\overset{−} {{j}}+\mathrm{1}}{\left(\overset{−} {{j}}−\mathrm{2}\right)\left(−\mathrm{2}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \\ $$$$=−\frac{\mathrm{2}\overset{−} {{j}}\:+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left(\overset{−} {{j}}−\mathrm{2}\right)}\:\:\Rightarrow{f}\left({x}\right)=\:\frac{\mathrm{5}}{\mathrm{7}\left({x}−\mathrm{2}\right)}\:+\frac{\mathrm{2}{j}+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({j}−\mathrm{2}\right)\left({x}−{j}\right)} \\ $$$$−\frac{\mathrm{2}\overset{−\:} {{j}}\:+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left(\overset{−} {{j}}−\mathrm{2}\right)\left({x}−\overset{−} {{j}}\right)}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{\mathrm{5}}{\mathrm{7}}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\mathrm{2}\right)^{{n}+\mathrm{1}} }\:\:+\frac{\mathrm{2}{j}+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({j}−\mathrm{2}\right)}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−{j}\right)^{{n}+\mathrm{1}} } \\ $$$$−\frac{\mathrm{2}\overset{−} {{j}}\:+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left(\overset{−} {{j}}−\mathrm{2}\right)}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\:{but} \\ $$$$\left.\mathrm{2}\right){f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\mathrm{5}\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{7}\left(−\mathrm{2}\right)^{{n}+\mathrm{1}} }\:+\frac{\mathrm{2}{j}+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({j}−\mathrm{2}\right)}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left(−{j}\right)^{{n}+\mathrm{1}} } \\ $$$$−\frac{\mathrm{2}\overset{−} {{j}}+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left(\overset{−} {{j}}−\mathrm{2}\right)}\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left(−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\: \\ $$$$=−\frac{\mathrm{5}}{\mathrm{7}}\:\frac{{n}!}{\mathrm{2}^{{n}+\mathrm{1}} }\:−\frac{\mathrm{2}{j}+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({j}−\mathrm{2}\right)}\:\frac{{n}!}{{j}^{{n}+\mathrm{1}} }\:+\frac{\mathrm{2}\overset{−} {{j}}\:+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left(\overset{−} {{j}}−\mathrm{2}\right)}\:\frac{{n}!}{\left(\overset{−} {{j}}\right)^{{n}+\mathrm{1}} } \\ $$$$\left.\mathrm{3}\right)\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left\{\:−\frac{\mathrm{5}}{\mathrm{7}\:.\mathrm{2}^{{n}+\mathrm{1}} }\:−\frac{\mathrm{2}{j}+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({j}−\mathrm{2}\right){j}^{{n}+\mathrm{1}} }\:+\frac{\mathrm{2}\overset{−} {{j}}\:+\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left(\overset{−} {{j}}−\mathrm{2}\right)\left(\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\right\}{x}^{{n}} \\ $$$$ \\ $$

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