Question Number 116365 by soumyasaha last updated on 03/Oct/20
Answered by Olaf last updated on 03/Oct/20
![f(x) = Σ_(k=0) ^3 ((f^((k)) (a))/(k!))(x−a)^k +R_3 (x) f^((0)) (a) = f((π/2)) = cos((π/2)) = 0 f^((1)) (a) = f^′ ((π/2)) = −sin((π/2)) = −1 f^((2)) (a) = f^′ ′((π/2)) = −cos((π/2)) = 0 f^((3)) (a) = f^′ ′′((π/2)) = sin((π/2)) = 1 f(x) = −(x−(π/2))+(((x−(π/2))^3 )/6)+R_3 (x) ∃ ξ∈[0;2π]╲(π/2), R_3 (x) = ((f^((4)) (ξ))/(4!))(x−(π/2))^4 ∣R_3 (x)∣ ≤ ((∣f^((4)) (ξ)∣)/(4!))(x−(π/2))^4 ≤ (1/(24))(x−90°)^4 For x = 80° : ∣R_3 (80°)∣ ≤ (1/(24))(10°)^4 = (1/(24))((π/(18)))^4 cos80° ≈ (π/(18))−(1/6)((π/(18)))^3 +(1/(24))((π/(18)))^4 cos80° ≈ 0,17369](https://www.tinkutara.com/question/Q116421.png)
$${f}\left({x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\frac{{f}^{\left({k}\right)} \left({a}\right)}{{k}!}\left({x}−{a}\right)^{{k}} +{R}_{\mathrm{3}} \left({x}\right) \\ $$$${f}^{\left(\mathrm{0}\right)} \left({a}\right)\:=\:{f}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$${f}^{\left(\mathrm{1}\right)} \left({a}\right)\:=\:{f}^{'} \left(\frac{\pi}{\mathrm{2}}\right)\:=\:−\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:−\mathrm{1} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({a}\right)\:=\:{f}^{'} '\left(\frac{\pi}{\mathrm{2}}\right)\:=\:−\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$${f}^{\left(\mathrm{3}\right)} \left({a}\right)\:=\:{f}^{'} ''\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{1} \\ $$$${f}\left({x}\right)\:=\:−\left({x}−\frac{\pi}{\mathrm{2}}\right)+\frac{\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{3}} }{\mathrm{6}}+{R}_{\mathrm{3}} \left({x}\right) \\ $$$$\exists\:\xi\in\left[\mathrm{0};\mathrm{2}\pi\right]\diagdown\frac{\pi}{\mathrm{2}},\:{R}_{\mathrm{3}} \left({x}\right)\:=\:\frac{{f}^{\left(\mathrm{4}\right)} \left(\xi\right)}{\mathrm{4}!}\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}} \\ $$$$\mid{R}_{\mathrm{3}} \left({x}\right)\mid\:\leqslant\:\frac{\mid{f}^{\left(\mathrm{4}\right)} \left(\xi\right)\mid}{\mathrm{4}!}\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{24}}\left({x}−\mathrm{90}°\right)^{\mathrm{4}} \\ $$$$\mathrm{For}\:{x}\:=\:\mathrm{80}°\:: \\ $$$$\mid{R}_{\mathrm{3}} \left(\mathrm{80}°\right)\mid\:\leqslant\:\frac{\mathrm{1}}{\mathrm{24}}\left(\mathrm{10}°\right)^{\mathrm{4}} \:=\:\frac{\mathrm{1}}{\mathrm{24}}\left(\frac{\pi}{\mathrm{18}}\right)^{\mathrm{4}} \\ $$$$\mathrm{cos80}°\:\approx\:\frac{\pi}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\pi}{\mathrm{18}}\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{24}}\left(\frac{\pi}{\mathrm{18}}\right)^{\mathrm{4}} \\ $$$$\mathrm{cos80}°\:\approx\:\mathrm{0},\mathrm{17369} \\ $$
Commented by soumyasaha last updated on 04/Oct/20
$$\mathrm{Thanks}\:\mathrm{Sir}. \\ $$