Question Number 33257 by prof Abdo imad last updated on 14/Apr/18
$${let}\:{g}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$\left.\mathrm{1}\right)\:{find}\:{g}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{if}\:{g}\left({x}\right)=\Sigma\:{u}_{{n}} \:{x}^{{n}} \:\:\:{find}\:{the}\:{sequence}\:{u}_{{n}} \\ $$
Commented by prof Abdo imad last updated on 15/Apr/18
$${let}\:{decompose}\:{g}\left({x}\right)\:{inside}\:{C}\left({x}\right) \\ $$$${g}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −{i}\right)\left({x}^{\mathrm{2}} +{i}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\left({x}−\sqrt{{i}}\right)\left({x}+\sqrt{{i}}\right)\left({x}\:−\sqrt{−{i}}\right)\left({x}+\sqrt{−{i}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\left({x}\:−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({x}\:+{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({x}\:\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\left(\:{x}\:+{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)} \\ $$$$=\:\frac{{a}}{{x}\:−{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:\:+\frac{{b}}{{x}\:+{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:+\:\frac{{c}}{{x}\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} }\:+\:\frac{{d}}{{x}\:+\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} } \\ $$$${a}\:=\:\frac{\mathrm{1}}{{p}^{'} \left(\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)}\:\:{with}\:{p}\left({x}\right)=\:\mathrm{1}+{x}^{\mathrm{4}} \:\Rightarrow{p}^{'} \left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} \\ $$$${if}\:{z}_{{k}} \:{pole}\:{p}^{'} \left({z}_{{k}} \right)\:=\:\mathrm{4}{z}_{{k}} ^{\mathrm{3}} \:\:=\frac{−\mathrm{4}}{{z}_{{k}} }\:\Rightarrow{a}\:=−\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}} \\ $$$${p}^{'} \left(−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:=\frac{−\mathrm{4}}{−{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:=\:\frac{\mathrm{4}}{{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:\Rightarrow\:{b}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$${p}^{'} \left(\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\:=\frac{−\mathrm{4}}{{e}^{−{i}\frac{\pi}{\mathrm{4}}} }\:\Rightarrow\:{c}\:=−\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${p}^{'} \:\left(\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\:\:=\:\:\frac{−\mathrm{4}}{−{e}^{−{i}\frac{\pi}{\mathrm{4}}} }\:\:\Rightarrow\:{d}=\:\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${g}\left({x}\right)\:=−\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}\left({x}\:−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)}\:\:+\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}\left(\:{x}\:+{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)}\:−\frac{{e}^{−{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}\left({x}\:−_{} {e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)} \\ $$$$+\:\frac{{e}^{−{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}\left(\:\:{x}\:+{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)}\:\Rightarrow \\ $$$${g}^{\left({n}\right)} \left({x}\right)\:=\:\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}}\:\left(\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}\:+{e}^{{i}\frac{\pi}{\mathrm{4}\left\{\right.}} \right)^{{n}+\mathrm{1}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}\:−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{{n}+\mathrm{1}} }\right) \\ $$$$+\:\frac{{e}^{−{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}}\:\left(\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}\:+\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)^{{n}+\mathrm{1}} }\:−\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \boldsymbol{{n}}!}{\left({x}\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)^{{n}+\mathrm{1}} }\right) \\ $$$${g}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!\:{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}}\left(\:\:{e}^{−{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{4}}} \:−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{e}^{−{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{4}}} \right) \\ $$$$+\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}}\left(\:\:{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{4}}} \:−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{4}}} \right)… \\ $$$${be}\:{continued}… \\ $$