Exercise 9
By definition 8,L(x) = sin π/6 + 2 sin π/6 cos π/6(x − π/6) = 1/2 + 1/√ 3(x − π/6).
Exercise 14
L(h) = g − R2g2h,the percentage of decreasing of gravity is R2h2 · 100%.
Exercise 17
Let f (x) = x1/4. Note f (81) = 3. This is the point we would like to ap- proximate f (85). For L(x) = f (81) + f0(81)(x − 81) = 3 + 1081 (x − 81), L(85) = 3 + 1084 . The error term can be obtained from second derivative.
f00(x) = −163 (x1/41 )7 < 0. Hence Error=|f00| < 361·16. The value would be contained in [3 + 1084 −Error, 3 + 1084 ].
Exercise 20
Let f (x) = sin x. Not f (π/6) = 1/2. For L(x) = f (π/6)+f0(π/6)(x−π/6) = 1/2 + √
3/2(x − π/6),L(π/5) = 1/2 +√
3/60π. Taking second derivative of f,f00(x) = − sin x, which is negative when x ∈ [π/6, π/5], the Error is
|f00| < sin π/4 = 1/√
2. The value would lie in [L(π/5)−Error,L(π/5)].
Exercise 27
By corollary C, in our case M = 0,N = 1/2, and f (3) ≈ 3 + 1/8, Error=
(N − M )/4(3 − 2)2 = 1/8. So the value f (3) is contained in [3, 3 + 1/4].
Exercise 28
By corollary C, in our case M = 1/6, N = 1/2 and f (3) ≈ 3 + 1/6, Error= 1/12. So the value f (3) is contained in [3 + 1/12, 3 + 1/4].
Exercise 29
L(x) = 1 + 2(x − 2), the approximation for g(1.8) is L(1.8) = 0.6. By Corollary B,|E(x)| < K/2(x − 2)2, in our case, K = 1 + (1.8 − 2)2 = 1.04.
So the maximum error would be |E(1.8)| < 1.04/2(1.04) = 1.0816/2.
1 LATEX
