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We shall use the figure and the notation given in π the text. We shall assume that the angle ϕ satisfies ϕ ∈ 0, . 2 ϕ below. The arc is denoted by , Figure 11: Circle of radius r, centre angle ϕ, thus periphery angle 2 and the corresponding cord is denoted by d. Finally, we let s denote the height on the dotted vertical diagonal. The approximating expression ˜ of the length is given in the form ˜ = ad + bs, where a and b are constants which will be found below. 1) First prove that ˜ = 2ar sin ϕ + br sin ϕ.

Com 48 Calculus Analyse 1c-6 Approximating polynomials I. 1) We get by successive differentiation f (x) = 6 sin x − 6x + x3 , f (x) = 6 cos x − 6 + 3x2 , f (x) = −6 sin x + 6x, f (3) (x) = −6 cos x + 6, f (4) (x) = 6 sin x, f (5) (x) = 6 cos x, f (0) = 0, f (0) = 6 − 6 + 0 = 0, f (0) = 0, f (3) (0) = 0, f (4) (0) = 0, f (5) (0) = 6. The searched Taylor polynomial is P5 (x) = 6 5 1 5 1 (5) f (0) x5 = x = x . 5! 120 20 2) We get by successive differentiation f (x) = ln(1 + x) − x, 1 f (x) = − 1, 1+x 1 , f (x) = − (1 + x)2 f (0) = 0, f (0) = 0, f (0) = −1.

From these results follows that a general term of the Taylor expansion is Q(j) (0) j 8! x = a8−j xj = j! (8 − j)! 8 j a8−j xj , j = 0, 1, . . , 8. Then we find the Taylor polynomial 8 8 j P8 (x) = j=0 a8−j xj . 2) Since Q(9) (x) ≡ 0, it follows from Taylor’s formula that Q(x) = (x + a)8 = P8 (x) + 1 (9) Q (ξ) x9 = P8 (x). 9! 3) When we use the same method as above we obtain the general binomial formula n (x + a)n = j=0 n j an−j xj . 3 Prove that for positive x, 1+ 1 3 1 6 x − x < 2 8 1 + x3 < 1 + 1 3 1 6 1 9 x − x + x , 2 8 16 and find the corresponding bounds for 1 2 1 + x3 dx.

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