let f(x)= (x/(x^2 +x+1)) 1) calculate f^((n)) (0) 2) developp f at integr serie .


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Question Number 36010 by abdo mathsup 649 cc last updated on 27/May/18

$${let}\:{f}\left({x}\right)=\:\:\frac{{x}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by prof Abdo imad last updated on 28/May/18
$let decompose f(x)inside C(x) the roots of p(x)=x^2 +x+1 are are j =e^(i((2π)/3)) and j^− =e^(−i((2π)/3)) so f(x)= (x/((x−j)(x−j^− ))) = (a/(x−j)) +(b/(x−j^− )) a=lim_(x→j) (x−j)f(x) = (j/(j−j^− )) = (j/(2 ((√3)/2))) = (j/(√3)) b =lim_(x→j^− ) (x−j^− )f(x)= (j^− /(j^− −j)) =−(j^− /(√3)) ⇒ f(x) = (j/((√3)(x−j))) −(j^− /((√3)(x−j^− ))) ⇒ f^((n)) (x) = (j/(√3)){ (((−1)^n n!)/((x−j)^(n+1) ))} −(j^− /(√3)) { (((−1)^n n!)/((x−j^− )^(n+1) ))} f^((n)) (x) = (((−1)^n n!)/(√3)){ (j/((x−j)^(n+1) )) −(j^− /((x−j^− )^(n+1) ))} =(((−1)^n n!)/(√3)){ ((j(x−j^− )^(n+1) −j^− (x−j)^(n+1) )/((x^2 +x+1)^(n+1) ))}$
$${let}\:{decompose}\:{f}\left({x}\right){inside}\:{C}\left({x}\right)\:{the}\:{roots}\:{of} \\ $$$${p}\left({x}\right)={x}^{\mathrm{2}} \:+{x}+\mathrm{1}\:{are}\:{are}\:{j}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:{and}\:\overset{−} {{j}}\:={e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:{so} \\ $$$${f}\left({x}\right)=\:\:\frac{{x}}{\left({x}−{j}\right)\left({x}−\overset{−} {{j}}\right)}\:=\:\frac{{a}}{{x}−{j}}\:+\frac{{b}}{{x}−\overset{−} {{j}}} \\ $$$${a}={lim}_{{x}\rightarrow{j}} \left({x}−{j}\right){f}\left({x}\right)\:=\:\frac{{j}}{{j}−\overset{−} {{j}}}\:=\:\frac{{j}}{\mathrm{2}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\:\frac{{j}}{\sqrt{\mathrm{3}}} \\ $$$${b}\:={lim}_{{x}\rightarrow\overset{−} {{j}}} \left({x}−\overset{−} {{j}}\right){f}\left({x}\right)=\:\frac{\overset{−} {{j}}}{\overset{−} {{j}}−{j}}\:=−\frac{\overset{−} {{j}}}{\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\frac{{j}}{\sqrt{\mathrm{3}}\left({x}−{j}\right)}\:−\frac{\overset{−} {{j}}}{\sqrt{\mathrm{3}}\left({x}−\overset{−} {{j}}\right)}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{{j}}{\sqrt{\mathrm{3}}}\left\{\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−{j}\right)^{{n}+\mathrm{1}} }\right\}\:−\frac{\overset{−} {{j}}}{\sqrt{\mathrm{3}}}\:\left\{\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\right\} \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\sqrt{\mathrm{3}}}\left\{\:\:\frac{{j}}{\left({x}−{j}\right)^{{n}+\mathrm{1}} }\:−\frac{\overset{−} {{j}}}{\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\sqrt{\mathrm{3}}}\left\{\:\frac{{j}\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} \:−\overset{−} {{j}}\left({x}−{j}\right)^{{n}+\mathrm{1}} }{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\right\} \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 28/May/18
$so f^((n)) (0) = (((−1)^n n!)/(√3)){ ((j(−j^− )^(n+1) −j^− (−j)^(n+1) )/1)} =(((−1)^n n!)/(√3)){ (−1)^(n+1) j(j^− )^(n+1) −(−1)^(n+1) j^− j^(n+1) } =(((−1)^n n!)/(√3)) (−1)^(n+1) { j^−^n −j^n } = (1/(√3)){ 2Im(j^n )) = (2/(√3)) sin(((2nπ)/3)) f^((n)) (0) = (2/(√3))sin(((2nπ)/3)). 2) we have f(x) = Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n ⇒ f(x) = Σ_(n=0) ^∞ (2/(√3)) ((sin(((2nπ)/3)))/(n!)) x^n . f(x) = (2/(√3)) Σ_(n=0) ^∞ ((sin(((2nπ)/3)))/(n!)) x^n .$
$${so}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\sqrt{\mathrm{3}}}\left\{\:\frac{{j}\left(−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} \:−\overset{−} {{j}}\left(−{j}\right)^{{n}+\mathrm{1}} }{\mathrm{1}}\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\sqrt{\mathrm{3}}}\left\{\:\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{j}\left(\overset{−} {{j}}\right)^{{n}+\mathrm{1}} \:−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \overset{−} {{j}}\:{j}^{{n}+\mathrm{1}} \right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\sqrt{\mathrm{3}}}\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left\{\:\:\overset{−^{{n}} } {{j}}\:\:−{j}^{{n}} \right\} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\left\{\:\:\mathrm{2}{Im}\left({j}^{{n}} \right)\right)\:=\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:{sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right) \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}{sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right). \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\:\frac{{sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)}{{n}!}\:{x}^{{n}} \:\:. \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{3}}\right)}{{n}!}\:{x}^{{n}} \:\:. \\ $$
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