Question Number 76190 by abdomathmax last updated on 25/Dec/19
$${let}\:{f}\left({x}\right)=\frac{{arctan}\left(\mathrm{1}+{x}\right)}{\mathrm{2}+{x}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
$$\left.\mathrm{1}\right)\:{f}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({arctan}\left(\mathrm{1}+{x}\right)\right)^{\left({k}\right)} ×\left(\frac{\mathrm{1}}{{x}+\mathrm{2}}\right)^{\left({n}−{k}\right)} \\ $$$${we}\:{have}\:\left({arctan}\left(\mathrm{1}+{x}\right)\right)^{\left(\mathrm{1}\right)} =\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}\:\Rightarrow \\ $$$$\left({arctan}\left({x}+\mathrm{1}\right)\right)^{\left({k}\right)} =\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}\right)^{\left({k}−\mathrm{1}\right)} \\ $$$$\Delta^{'} =\mathrm{1}−\mathrm{2}\:=−\mathrm{1}\:\Rightarrow{x}_{\mathrm{1}} =−\mathrm{1}+{i}\:\:{and}\:{x}_{\mathrm{2}} =−\mathrm{1}−{i} \\ $$$${x}_{\mathrm{1}} =\sqrt{\mathrm{2}}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:=\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \:\:{and}\:{x}_{\mathrm{2}} =\sqrt{\mathrm{2}}{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}\:=\frac{\mathrm{1}}{\left({x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)\left({x}−\sqrt{\mathrm{2}}{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \right)}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{2}{isin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\left(\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{2}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\left\{\:\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} }\right\}\:\Rightarrow \\ $$$$\frac{{d}^{{k}} }{{dx}^{{k}} }\left({arctan}\left({x}+\mathrm{1}\right)\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} }−\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\frac{\left({x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} −\left({x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)^{{k}} }\right\} \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:={arctan}\left({x}+\mathrm{1}\right)\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+\mathrm{2}\right)^{{n}+\mathrm{1}} } \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!×\frac{\left({x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} −\left({x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)^{{k}} }×\frac{\left(−\mathrm{1}\right)^{{n}−{k}} \left({n}−{k}\right)!}{\left({x}+\mathrm{2}\right)^{{n}−{k}+\mathrm{1}} } \\ $$$$=\left(−\mathrm{1}\right)^{{n}} {n}!\:\frac{{arctan}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{2}\right)^{{n}+\mathrm{1}} } \\ $$$$+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({k}−\mathrm{1}\right)!\frac{{Im}\left\{\left({x}−\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} \right\}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)^{{k}} }×\frac{\left({n}−{k}\right)!}{\left({x}+\mathrm{2}\right)^{{n}−{k}+\mathrm{1}} } \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\pi}{\mathrm{4}}\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$$+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\frac{\left({k}−\mathrm{1}\right)!\left({n}−{k}\right)!}{\mathrm{2}^{{k}} ×\mathrm{2}^{{n}−{k}+\mathrm{1}} }×{Im}\left\{\left(−\sqrt{\mathrm{2}}\right)^{{k}} \:{e}^{{i}\frac{\mathrm{3}{k}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\pi}{\mathrm{4}}\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}^{{n}+\mathrm{1}} }\:+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{n}!}{{k}!\left({n}−{k}\right)!}×\frac{\left({k}−\mathrm{1}\right)!\left({n}−{k}!\right.}{\mathrm{2}^{{n}+\mathrm{1}} }×\left(−\sqrt{\mathrm{2}}\right)^{{k}} {sin}\left(\frac{\mathrm{3}{k}\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\pi}{\mathrm{4}}×\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}^{{n}+\mathrm{1}} }\:+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}^{{n}+\mathrm{1}} }\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{n}!}{{k}}\left(−\sqrt{\mathrm{2}}\right)^{{k}} \:{sin}\left(\frac{\mathrm{3}{k}\pi}{\mathrm{4}}\right) \\ $$