Question Number 62440 by mathsolverby Abdo last updated on 21/Jun/19
$${let}\:{h}\left({x}\right)=\:{arctan}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\left.\mathrm{1}\right){calculate}\:{h}^{\left({n}\right)} \left({x}\right)\:{and}\:{h}^{\left({n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\left({x}\right){at}\:{integr}\:{serie}\:{at}\:{x}_{\mathrm{0}} =\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 23/Jun/19
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{h}^{'} \left({x}\right)=\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\mathrm{1}+\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}\:=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${x}^{\mathrm{2}} ={t}\:\Rightarrow{h}^{'} \left({x}\right)\:={g}\left({t}\right)\:=\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{3}{t}\:+\mathrm{1}} \\ $$$$\Delta\:=\mathrm{9}−\mathrm{4}\:=\mathrm{5}\:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow{g}\left({t}\right)\:=\frac{{t}−\mathrm{1}}{\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)} \\ $$$$=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −\frac{−\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)}\:=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} +\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)} \\ $$$$=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left\{\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\right\}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left\{\:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}\:−\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left\{\:\mathrm{1}−\frac{\mathrm{1}+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}\:−\left(\mathrm{1}−\frac{\mathrm{1}+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\right)\right\} \\ $$$$=\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\left\{\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}\right\}\:+\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\left\{\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\right\}\:{let}\:\:\alpha\:=\sqrt{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}}\:\:{and}\:\beta\:=\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$$\Rightarrow{h}^{'} \left({x}\right)=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\left\{\:\:\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\right\} \\ $$$$=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\left\{\:\:\frac{\mathrm{1}}{\left({x}+{i}\alpha\right)\left({x}−{i}\alpha\right)}\:+\frac{\mathrm{1}}{\left({x}−{i}\beta\right)\left({x}+{i}\beta\right)}\right\} \\ $$$$=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\left\{\:\:\frac{\mathrm{1}}{\mathrm{2}{i}\alpha}\:\left(\frac{\mathrm{1}}{{x}−{i}\alpha}\:−\frac{\mathrm{1}}{{x}+{i}\alpha}\right)\:+\frac{\mathrm{1}}{\mathrm{2}{i}\beta}\left(\:\frac{\mathrm{1}}{{x}−{i}\beta}\:−\frac{\mathrm{1}}{{x}+{i}\beta}\right)\right\}\Rightarrow \\ $$$${h}^{\left({n}\right)} \left({x}\right)\:=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}{i}\alpha}\left\{\:\left(\frac{\mathrm{1}}{{x}−{i}\alpha}\right)^{\left({n}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{{x}+{i}\alpha}\right)^{{n}−\mathrm{1}} \right\}+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}{i}\beta}\left\{\:\left(\frac{\mathrm{1}}{{x}−{i}\beta}\right)^{\left({n}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{{x}+{i}\beta}\right)^{\left({n}−\mathrm{1}\right)} \right\} \\ $$$$=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}{i}\alpha}\left\{\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}−{i}\alpha\right)^{{n}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+{i}\alpha\right)^{{n}} }\right\} \\ $$$$+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}{i}\beta}\left\{\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}−{i}\beta\right)^{{n}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+{i}\beta\right)^{{n}} }\right\} \\ $$$$=\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{4}{i}}\left\{\frac{\mathrm{1}}{\alpha}\left(\:\:\frac{\mathrm{1}}{\left({x}−{i}\alpha\right)^{{n}} }\:−\frac{\mathrm{1}}{\left({x}+{i}\alpha\right)^{{n}} }\right)\:+\frac{\mathrm{1}}{\beta}\left(\frac{\mathrm{1}}{\left({x}−{i}\beta\right)^{{n}} }\:−\frac{\mathrm{1}}{\left({x}+{i}\beta\right)^{{n}} }\right)\right\} \\ $$$$=\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{4}{i}}\left\{\frac{\mathrm{2}{i}}{\alpha}\:\frac{{Im}\left({x}+{i}\alpha\right)^{{n}} }{\left({x}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} \right)^{{n}} }\:+\frac{\mathrm{2}{i}}{\beta}\:\frac{{Im}\left({x}+{i}\beta\right)^{{n}} }{\left({x}^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)^{{n}} }\right\} \\ $$$${h}^{\left({n}\right)} \left({x}\right)=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left\{\:\frac{{Im}\left({x}+{i}\alpha\right)^{{n}} }{\alpha\left({x}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} \right)^{{n}} }\:+\frac{{Im}\left({x}+{i}\beta\right)^{{n}} }{\left({x}^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)^{{n}} }\right\} \\ $$$${h}^{\left({n}\right)} \left(\mathrm{1}\right)\:=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left\{\:\frac{{Im}\left(\mathrm{1}+{i}\alpha\right)^{{n}} }{\alpha\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)^{{n}} }\:+\frac{{Im}\left(\mathrm{1}+{i}\beta\right)^{{n}} }{\left(\mathrm{1}+\beta^{\mathrm{2}} \right)^{{n}} }\right\}\:. \\ $$$$\left.\mathrm{3}\right){h}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{h}^{\left({n}\right)} \left(\mathrm{1}\right)}{{n}!}\left({x}−\mathrm{1}\right)^{{n}} \:\:\:\:{and}\:{h}^{\left({n}\right)} \left(\mathrm{1}\right)\:{is}\:{known}. \\ $$
Commented by mathmax by abdo last updated on 23/Jun/19
$${error}\:{of}\:{typo}\:\:\: \\ $$$${h}^{\left({n}\right)} \left({x}\right)\:=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left\{\:\frac{{Im}\left({x}+{i}\alpha\right)^{{n}} }{\alpha\left({x}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} \right)^{{n}} }\:+\frac{{Im}\left({x}+{i}\beta\right)^{{n}} }{\beta\left({x}^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)^{{n}} }\right\}. \\ $$