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Question Number 31546 by prof Abdo imad last updated on 10/Mar/18

let consider the numrtical function f(x)= (1/(x^2 +x+1)) calculate f^((n)) (x) then give f^((n)) (0).

$${let}\:{consider}\:{the}\:{numrtical}\:{function} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{then}\:{give} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right). \\ $$

Answered by prof Abdo imad last updated on 10/Apr/18

let decompose f(x) inside C(x) roots of x^2 +x +1 =0 Δ=1−4=−3=(i(√3))^2 ⇒x_1 =((−1 +i(√3))/2) =j x_2 =((−1−i(√3))/2) =j^− so f(x) = (1/((x−j)(x−j^− ))) = (a/(x−j)) +(b/(x−j^− )) a= (1/(j−j^− )) = (1/(i(√3))) ,b = (1/(j^− −j)) = (1/(−i(√3))) ⇒ f(x) = (1/(i(√3)))( (1/(x−j)) − (1/(x−j^− ))) ⇒ f^((n)) (x)= (1/(i(√3)))( ((1/(x−j)))^n −((1/(x−j^− )))^n ) = (1/(i(√3)))( (((−1)^n n!)/((x−j)^(n+1) )) −(((−1)^n n!)/((x−j^− )^(n+1) ))) = (((−1)^n n!)/(i(√3)))( (1/((x−j)^(n+1) )) − (1/((x−j^− )^(n+1) ))) ⇒ f^((n)) (0) = (((−1)^n n!)/(i(√3))) ( (((−1)^(n+1) )/j^(n+1) ) −(((−1)^(n+1) )/j^−^(n+1) )) = ((−n!)/(i(√3)))( ((−j^(n+1) +j^(−n+1) )/1))= ((n!)/(i(√3)))( j^(n+1) −j^−^(n+1) ) = ((n!)/(i(√3)))( 2i Im(j^(n+1) ) = ((2n!)/(√3)) Im( e^(2i(n+1)(π/3)) ) ⇒ f^((n)) (0) = ((2(n!))/(√3)) sin(((2(n+1)π)/3)) .

$${let}\:{decompose}\:{f}\left({x}\right)\:{inside}\:{C}\left({x}\right) \\ $$$${roots}\:{of}\:{x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}\:+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={j} \\ $$$${x}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\overset{−} {{j}}\:\:{so} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left({x}−{j}\right)\left({x}−\overset{−} {{j}}\right)}\:=\:\frac{{a}}{{x}−{j}}\:\:+\frac{{b}}{{x}−\overset{−} {{j}}} \\ $$$${a}=\:\frac{\mathrm{1}}{{j}−\overset{−} {{j}}}\:=\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\:,{b}\:=\:\frac{\mathrm{1}}{\overset{−} {{j}}\:−{j}}\:=\:\frac{\mathrm{1}}{−{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\left(\:\:\frac{\mathrm{1}}{{x}−{j}}\:−\:\frac{\mathrm{1}}{{x}−\overset{−} {{j}}}\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\left(\:\:\left(\frac{\mathrm{1}}{{x}−{j}}\right)^{{n}} \:−\left(\frac{\mathrm{1}}{{x}−\overset{−} {{j}}}\right)^{{n}} \right) \\ $$$$=\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\left(\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left({x}−{j}\right)^{{n}+\mathrm{1}} }\:\:−\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\right) \\ $$$$=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}\sqrt{\mathrm{3}}}\left(\:\:\:\frac{\mathrm{1}}{\left({x}−{j}\right)^{{n}+\mathrm{1}} }\:−\:\frac{\mathrm{1}}{\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}\sqrt{\mathrm{3}}}\:\left(\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{j}^{{n}+\mathrm{1}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\overset{−^{{n}+\mathrm{1}} } {{j}}}\right) \\ $$$$=\:\frac{−{n}!}{{i}\sqrt{\mathrm{3}}}\left(\:\:\frac{−{j}^{{n}+\mathrm{1}} \:+\overset{−{n}+\mathrm{1}} {{j}}}{\mathrm{1}}\right)=\:\frac{{n}!}{{i}\sqrt{\mathrm{3}}}\left(\:{j}^{{n}+\mathrm{1}} \:−\overset{−^{{n}+\mathrm{1}} } {{j}}\right) \\ $$$$=\:\:\frac{{n}!}{{i}\sqrt{\mathrm{3}}}\left(\:\mathrm{2}{i}\:{Im}\left({j}^{{n}+\mathrm{1}} \right)\:=\:\frac{\mathrm{2}{n}!}{\sqrt{\mathrm{3}}}\:\:{Im}\left(\:\:{e}^{\mathrm{2}{i}\left({n}+\mathrm{1}\right)\frac{\pi}{\mathrm{3}}} \right)\:\Rightarrow\right. \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\mathrm{2}\left({n}!\right)}{\sqrt{\mathrm{3}}}\:{sin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)\:. \\ $$$$ \\ $$

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