Question Number 31546 by prof Abdo imad last updated on 10/Mar/18
$${let}\:{consider}\:{the}\:{numrtical}\:{function} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{then}\:{give} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right). \\ $$
Answered by prof Abdo imad last updated on 10/Apr/18
$${let}\:{decompose}\:{f}\left({x}\right)\:{inside}\:{C}\left({x}\right) \\ $$$${roots}\:{of}\:{x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}=\left({i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}\:+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:={j} \\ $$$${x}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\overset{−} {{j}}\:\:{so} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left({x}−{j}\right)\left({x}−\overset{−} {{j}}\right)}\:=\:\frac{{a}}{{x}−{j}}\:\:+\frac{{b}}{{x}−\overset{−} {{j}}} \\ $$$${a}=\:\frac{\mathrm{1}}{{j}−\overset{−} {{j}}}\:=\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\:,{b}\:=\:\frac{\mathrm{1}}{\overset{−} {{j}}\:−{j}}\:=\:\frac{\mathrm{1}}{−{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\left(\:\:\frac{\mathrm{1}}{{x}−{j}}\:−\:\frac{\mathrm{1}}{{x}−\overset{−} {{j}}}\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\left(\:\:\left(\frac{\mathrm{1}}{{x}−{j}}\right)^{{n}} \:−\left(\frac{\mathrm{1}}{{x}−\overset{−} {{j}}}\right)^{{n}} \right) \\ $$$$=\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{3}}}\left(\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left({x}−{j}\right)^{{n}+\mathrm{1}} }\:\:−\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\right) \\ $$$$=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}\sqrt{\mathrm{3}}}\left(\:\:\:\frac{\mathrm{1}}{\left({x}−{j}\right)^{{n}+\mathrm{1}} }\:−\:\frac{\mathrm{1}}{\left({x}−\overset{−} {{j}}\right)^{{n}+\mathrm{1}} }\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}\sqrt{\mathrm{3}}}\:\left(\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{j}^{{n}+\mathrm{1}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\overset{−^{{n}+\mathrm{1}} } {{j}}}\right) \\ $$$$=\:\frac{−{n}!}{{i}\sqrt{\mathrm{3}}}\left(\:\:\frac{−{j}^{{n}+\mathrm{1}} \:+\overset{−{n}+\mathrm{1}} {{j}}}{\mathrm{1}}\right)=\:\frac{{n}!}{{i}\sqrt{\mathrm{3}}}\left(\:{j}^{{n}+\mathrm{1}} \:−\overset{−^{{n}+\mathrm{1}} } {{j}}\right) \\ $$$$=\:\:\frac{{n}!}{{i}\sqrt{\mathrm{3}}}\left(\:\mathrm{2}{i}\:{Im}\left({j}^{{n}+\mathrm{1}} \right)\:=\:\frac{\mathrm{2}{n}!}{\:\sqrt{\mathrm{3}}}\:\:{Im}\left(\:\:{e}^{\mathrm{2}{i}\left({n}+\mathrm{1}\right)\frac{\pi}{\mathrm{3}}} \right)\:\Rightarrow\right. \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\frac{\mathrm{2}\left({n}!\right)}{\:\sqrt{\mathrm{3}}}\:{sin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{\mathrm{3}}\right)\:. \\ $$$$ \\ $$