112 4.5 Taylor’s Theorem Remark 4.5.1 The conclusion of Taylor’s theorem still holds true if x =x0. In this case, c =x =x0. ■ Example 4.5.1 We will use Taylor’s theorem to estimate the error in approximating the function f (x) =sinxwith its 3rd Taylor polynomial at x0 =0 on the interval [−π/2,π/2]. Since f ′(x) =cosx, f ′′(x) =−sinx and f ′′′(x) =−cosx, a direct calculation shows that P3(x) =x− x3 3! . Moreover, for any c ∈Rwe have | f (4)(c)| =|sinc| ≤1. Therefore, for x ∈[−π/2,π/2] we get (for some c between x and 0), |sinx−P3(x)| =| f (4)(c)| 4! | x| ≤ π/2 4! ≤ 0.066. Theorem 4.5.2 Let nbe an even positive integer. Suppose f (n) exists and continuous on (a,b). Let x0 ∈(a,b) satisfy f ′(x0) =. . . =f (n−1)(x 0) =0 and f (n)(x 0)̸ =0. The following hold: (i) f (n)(x 0) >0 if and only if f has a local minimum at x0. (ii) f (n)(x 0) <0 if and only if f has a local maximum at x0. Proof: We will prove (i). Suppose f (n)(x 0) >0. Since f (n)(x 0) >0 and f (n) is continuous at x 0, there exists δ >0 such that f (n)(t) >0 for all t ∈(x 0 −δ,x0 +δ) ⊂(a,b). Fix anyx ∈(x0 −δ,x0 +δ). By Taylor’s theorem (Theorem 4.5.1) and the given assumption, there exists c in betweenx0 andx such that f (x) =f (x0)+ f (n)(c) n! (x−x0) n. Since nis even and c ∈(x0 −δ,x0 +δ), we have f (x) ≥f (x0). Thus, f has a local minimum at x0. Now, for the converse, suppose that f has a local minimum at x0. Then there exists δ >0 such that f (x) ≥f (x0) for all x ∈(x0 −δ,x0 +δ) ⊂(a,b). Fix a sequence {xk} in (a,b) that converges to x0 with xk̸ =x0 for every k. By Taylor’s theorem (Theorem 4.5.1), there exists a sequence {ck}, withck between xk andx0 for each k, such that f (xk) =f (x0)+ f (n)(c k) n! (xk −x0) n. Since xk ∈(x0 −δ,x0 +δ) for sufficiently large k, we have f (xk) ≥f (x0)
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