Question Number 30753 by abdo imad last updated on 25/Feb/18
$${let}\:{f}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{{x}^{{n}+\mathrm{1}} }\:\left({e}^{{x}} \:\:−\sum_{{p}=\mathrm{0}} ^{{n}\:} \:\:\:\frac{{x}^{{p}} }{{p}!}\right) \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{f}^{\left({n}\right)} \left({x}\right)=\frac{{Q}_{{n}} \left({x}\right)\:{e}^{{x}} \:−{P}_{{n}} \left({x}\right)}{{x}^{\mathrm{2}{n}+\mathrm{1}} }\:{find}\:{the} \\ $$$${polynomial}\:{P}_{{n}} \:{and}\:{Q}_{{n}} . \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{e}^{{x}} \:−\frac{{P}_{{n}} \left({x}\right)}{{Q}_{{n}} \left({x}\right)}={o}\left({x}^{\mathrm{2}{n}+\mathrm{1}} \:\:\right) \\ $$