Question Number 116365 by soumyasaha last updated on 03/Oct/20
Answered by Olaf last updated on 03/Oct/20
$${f}\left({x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\frac{{f}^{\left({k}\right)} \left({a}\right)}{{k}!}\left({x}−{a}\right)^{{k}} +{R}_{\mathrm{3}} \left({x}\right) \\ $$$${f}^{\left(\mathrm{0}\right)} \left({a}\right)\:=\:{f}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$${f}^{\left(\mathrm{1}\right)} \left({a}\right)\:=\:{f}^{'} \left(\frac{\pi}{\mathrm{2}}\right)\:=\:−\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:−\mathrm{1} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({a}\right)\:=\:{f}^{'} '\left(\frac{\pi}{\mathrm{2}}\right)\:=\:−\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$${f}^{\left(\mathrm{3}\right)} \left({a}\right)\:=\:{f}^{'} ''\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{1} \\ $$$${f}\left({x}\right)\:=\:−\left({x}−\frac{\pi}{\mathrm{2}}\right)+\frac{\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{3}} }{\mathrm{6}}+{R}_{\mathrm{3}} \left({x}\right) \\ $$$$\exists\:\xi\in\left[\mathrm{0};\mathrm{2}\pi\right]\diagdown\frac{\pi}{\mathrm{2}},\:{R}_{\mathrm{3}} \left({x}\right)\:=\:\frac{{f}^{\left(\mathrm{4}\right)} \left(\xi\right)}{\mathrm{4}!}\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}} \\ $$$$\mid{R}_{\mathrm{3}} \left({x}\right)\mid\:\leqslant\:\frac{\mid{f}^{\left(\mathrm{4}\right)} \left(\xi\right)\mid}{\mathrm{4}!}\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{24}}\left({x}−\mathrm{90}°\right)^{\mathrm{4}} \\ $$$$\mathrm{For}\:{x}\:=\:\mathrm{80}°\:: \\ $$$$\mid{R}_{\mathrm{3}} \left(\mathrm{80}°\right)\mid\:\leqslant\:\frac{\mathrm{1}}{\mathrm{24}}\left(\mathrm{10}°\right)^{\mathrm{4}} \:=\:\frac{\mathrm{1}}{\mathrm{24}}\left(\frac{\pi}{\mathrm{18}}\right)^{\mathrm{4}} \\ $$$$\mathrm{cos80}°\:\approx\:\frac{\pi}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\pi}{\mathrm{18}}\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{24}}\left(\frac{\pi}{\mathrm{18}}\right)^{\mathrm{4}} \\ $$$$\mathrm{cos80}°\:\approx\:\mathrm{0},\mathrm{17369} \\ $$
Commented by soumyasaha last updated on 04/Oct/20
$$\mathrm{Thanks}\:\mathrm{Sir}. \\ $$