- #1
lep11
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Let ##f:\mathbb{R^2} \rightarrow\mathbb{R}, f(x)=\cos(x_1^2+x_2^3)## and ##f(0,3;-0,3)\approx1.##Show that the error of that approximationis less than ##7\cdot10^{-3}.##
Obviously ##\mathbb{R^2}## is convex, that is, any points ##a,b\in\mathbb{R^2}## can be connected with a line segment. In addition, ##f## is differentiable as a composition of two differentiable functions. Thus, the conditions of mean value theorem for vector functions are satisfied. By applying the theorem we get
$$|\cos(0,3^2+(-0,3)^3)-1|$$
$$=|\cos(0,3^2-0,3^3)-\cos(0^2+0^3)|$$
$$=|Df(c)((0,3;-0,3)-(0,0))|$$
$$=|<\nabla{f(c)},(0,3;-0,3)>|$$
$$\leq\|\nabla{f(c)}\|\|(0,3;-0,3)\|$$(by C-S inequality)
$$=\|\nabla{f(c)}\|\sqrt{0,3^2+0,3^2},$$
$$=\sqrt{(-2c_1\sin(c_1^2+c_2^3)^2+(-3c_2^2\sin(c_1^2+c_2^3))^2}\sqrt{0,3^2+0,3^2},$$
$$\leq{\sqrt{(4c_1^2+9c_2^4}\sqrt{0,3^2+0,3^2}}$$
$$\leq...?$$
where ##c=(c_1,c_2)## lies on the line connecting ##(0,3;-0,3)## and the origin. Consequently ##\|c\|=\sqrt{c_1^2+c_2^2}\leq\sqrt{0,3^2+0,3^2}.## Further, ##\nabla{f(c)}=(-2c_1\sin(c_1^2+c_2^3),-3c_2^2\sin(c_1^2+c_2^3)).##
How could one estimate ##\|\nabla{f(c)}\|?## It is a matrix (Frobenius) norm, right?
How to proceed?
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