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let-g-x-x-1-x-2-x-1-1-find-g-n-x-2-calculate-g-n-0-3-developp-g-at-integr-serie-

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Question Number 42688 by prof Abdo imad last updated on 31/Aug/18
let g(x) =((x−1)/(x^2 +x +1)) 1) find g^((n)) (x) 2)calculate g^((n)) (0) 3) developp g at integr serie.
$${let}\:{g}\left({x}\right)\:=\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}\:+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:\:{find}\:{g}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{g}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{g}\:{at}\:\:{integr}\:{serie}. \\ $$
Commented by maxmathsup by imad last updated on 02/Sep/18
1) poles of g? roots of x^2 +x+1 are j=e^(i((2π)/3)) and j^− =e^(−((i2π)/3)) g(x)=((x−1)/((x−j)(x−j^− ))) =(a/(x−j)) +(b/(x−j^− )) a =lim_(x→j) (x−j)g(x) =((j−1)/(j−j^− )) =((j−1)/(2 i((√3)/2))) =((j−1)/(i(√3))) b =lim_(x→j^− ) (x−j^− )g(x) =((j^− −1)/((j^− −j))) =((j^− −1)/(−i(√3))) =((1−j^− )/(i(√3))) ⇒ g(x) = ((j−1)/(i(√3)(x−j))) −((j^− −1)/(i(√3)(x−j^− ))) ⇒ g^((n)) (x)=((j−1)/(i(√3))) (((−1)^n n!)/((x−j)^(n+1) )) −((j^− −1)/(i(√3))) (((−1)^n n!)/((x−j^− )^(n+1) )) =(((−1)^n n!)/(i(√3))){ ((j−1)/((x−j)^(n+1) )) −((j^− −1)/((x−j^− )^(n+1) ))} =(((−1)^n n!)/(i(√3))){ (((j−1)(x−j^− )^(n+1) −(j^− −1)(x−j)^(n+1) )/((x^2 +x+1)^(n+1) ))}
$$\left.\mathrm{1}\right)\:{poles}\:{of}\:{g}?\:\:\:\:\:\:{roots}\:{of}\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:\:{are}\:{j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:\:{and}\:\overset{−} {{j}}={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${g}\left({x}\right)=\frac{{x}−\mathrm{1}}{\left({x}−{j}\right)\left({x}−\overset{−} {{j}}\right)}\:=\frac{{a}}{{x}−{j}}\:+\frac{{b}}{{x}−\overset{−} {{j}}} \\ $$$${a}\:={lim}_{{x}\rightarrow{j}} \left({x}−{j}\right){g}\left({x}\right)\:=\frac{{j}−\mathrm{1}}{{j}−\overset{−} {{j}}}\:=\frac{{j}−\mathrm{1}}{\mathrm{2}\:{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{{j}−\mathrm{1}}{{i}\sqrt{\mathrm{3}}} \\ $$$${b}\:={lim}_{{x}\rightarrow\overset{−} {{j}}} \:\left({x}−\overset{−} {{j}}\right){g}\left({x}\right)\:=\frac{\overset{−} {{j}}−\mathrm{1}}{\left(\overset{−} {{j}}−{j}\right)}\:=\frac{\overset{−} {{j}}−\mathrm{1}}{−{i}\sqrt{\mathrm{3}}}\:=\frac{\mathrm{1}−\overset{−} {{j}}}{{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${g}\left({x}\right)\:=\:\frac{{j}−\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({x}−{j}\right)}\:\:−\frac{\overset{−} {{j}}−\mathrm{1}}{{i}\sqrt{\mathrm{3}}\left({x}−\overset{−} {{j}}\right)}\:\Rightarrow \\ $$$${g}^{\left({n}\right)} \left({x}\right)=\frac{{j}−\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−{j}\right)^{{n}+\mathrm{1}} }\:−\frac{\overset{−} {{j}}−\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} } \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}\sqrt{\mathrm{3}}}\left\{\:\frac{{j}−\mathrm{1}}{\left({x}−{j}\right)^{{n}+\mathrm{1}} }\:−\frac{\overset{−} {{j}}−\mathrm{1}}{\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}\sqrt{\mathrm{3}}}\left\{\:\frac{\left({j}−\mathrm{1}\right)\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} \:−\left(\overset{−} {{j}}−\mathrm{1}\right)\left({x}−{j}\right)^{{n}+\mathrm{1}} }{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\right\} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 02/Sep/18
2) g^((n)) (0) = (((−1)^n n!)/(i(√3))){ (j−1)(−j^− )^(n+1) −(j^− −1)(−j)^(n+1) } =−((n!)/(i(√3))){ (j−1)(j^− )^(n+1) −(j^− −1)j^(n+1) } =((n!)/(i(√3))) 2iIm((j^− −1)(j)^(n+1) ) but (j^− −1)j^(n+1) =( e^(−((i2π)/3)) −1)e^(i((2(n+1)π)/3)) =(−(1/2) −i((√3)/2))(cos(((2(n+1)π)/3))+i sin(((2(n+1)π)/3))) =−(1/2) cos(((2(n+1)π)/3))−(i/2) sin(((2(n+1)π)/3))−((i(√3))/2) cos(((2(n+1)π)/3)) +((√3)/2) cos(((2(n+1)π)/3))sin(((2(n+1)π)/3)) ⇒ g^((n)) (0) =((2n!)/( (√3))) {−(1/2) sin(((2(n+1)π)/3))−((√3)/2) cos(((2(n+1)π)/3))} = −((n!)/( (√3))){ sin(((2(n+1)π)/3)) +(√3)cos(((2(n+1)π)/3)) .
$$\left.\mathrm{2}\right)\:{g}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}\sqrt{\mathrm{3}}}\left\{\:\left({j}−\mathrm{1}\right)\left(−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} \:−\left(\overset{−} {{j}}−\mathrm{1}\right)\left(−{j}\right)^{{n}+\mathrm{1}} \right\} \\ $$$$=−\frac{{n}!}{{i}\sqrt{\mathrm{3}}}\left\{\:\:\left({j}−\mathrm{1}\right)\left(\overset{−} {{j}}\right)^{{n}+\mathrm{1}} \:−\left(\overset{−} {{j}}−\mathrm{1}\right){j}^{{n}+\mathrm{1}} \right\}\:=\frac{{n}!}{{i}\sqrt{\mathrm{3}}}\:\mathrm{2}{iIm}\left(\left(\overset{−} {{j}}−\mathrm{1}\right)\left({j}\right)^{{n}+\mathrm{1}} \right)\:{but} \\ $$$$\left(\overset{−} {{j}}−\mathrm{1}\right){j}^{{n}+\mathrm{1}} \:=\left(\:{e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} −\mathrm{1}\right){e}^{{i}\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}} \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{2}}\:−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left({cos}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)+{i}\:{sin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:{cos}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)−\frac{{i}}{\mathrm{2}}\:{sin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)−\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{cos}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right) \\ $$$$+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{cos}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right){sin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)\:\Rightarrow \\ $$$${g}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\mathrm{2}{n}!}{\:\sqrt{\mathrm{3}}}\:\left\{−\frac{\mathrm{1}}{\mathrm{2}}\:{sin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{cos}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)\right\} \\ $$$$=\:−\frac{{n}!}{\:\sqrt{\mathrm{3}}}\left\{\:\:{sin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)\:+\sqrt{\mathrm{3}}{cos}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)\:.\right. \\ $$
Commented by maxmathsup by imad last updated on 02/Sep/18
3) we have g(x) =Σ_(n=0) ^∞ ((g^((n)) (0))/(n!)) x^n ⇒ g(x) =−(1/( (√3))) Σ_(n=0) ^∞ { sin(((2(n+1)π)/3)) +(√3)cos(((2(n+1)π)/3))}x^n .
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{g}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{g}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:\Rightarrow \\ $$$${g}\left({x}\right)\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\left\{\:{sin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)\:+\sqrt{\mathrm{3}}{cos}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)\right\}{x}^{{n}} \:. \\ $$

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