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)= sinx with 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 3x P3(x)= x − . 3! Moreover, for any c ∈ R we have | f (4)(c)| = |sinc|≤ 1. Therefore, for x ∈ [−π/2,π/2] we get (for some c between x and 0), | f (4)(c)| π/2 |sinx − P3(x)| = |x|≤ ≤ 0.066. 4! 4! Theorem 4.5.2 Let n be 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 any x ∈ (x0 − δ ,x0 + δ ). By Taylor’s theorem (Theorem 4.5.1) and the given assumption, there exists c in between x0 and x such that f (n)(c) f (x)= f (x0)+ (x − x0) n . n! Since n is 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 xk ̸ = x0 for every k. By Taylor’s theorem (Theorem 4.5.1), there exists a sequence {ck}, with ck between xk and x0 for each k, such that f (n)(c k) f (xk)= f (x0)+ (xk − x0) n . n! Since xk ∈ (x0 − δ ,x0 + δ ) for suffciently large k, we have f (xk) ≥ f (x0)
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