113 for such k. It follows that f (n)(c k) f (xk) − f (x0)= (xk − x0) n ≥ 0. n! This implies f (n)(c k) ≥ 0 for such k. Since {ck} converges to x0, f (n)(x 0)= limk →∞ f (n)(c k) ≥ 0. The proof of (ii) is similar. □ 2 ■ Example 4.5.2 Consider the function f (x)= x cosx defned on R. Then f ′ (x)= 2xcosx−x 2 sinx 2 and f ′′ (x)= 2cosx− 4xsinx − x cosx. Then f (0)= f ′ (0)= 0 and f ′′ (0)= 2 > 0. It follows from the previous theorem that f has a local minimum at 0. Notice, by the way, that since f (0)= 0 and f (π) < 0, 0 is not a global minimum. 6 5 3 2 ■ Example 4.5.3 Consider the function f (x)= −x + 2x + x 4 − 4x + x + 2x − 3 defned on R. A ′′ (1)= ′′′ (1)= direct calculations shows f ′ (1)= f f f (4)(1)= 0 and f (5)(1) < 0. It follows from the previous theorem that f has a local maximum at 1. Exercises 4.5.1 ▷ Use Taylor’s theorem to prove that for all x > 0 and m ∈ N, m kx xe > ∑ . k! k=0 4.5.2 Find the 5th Taylor polynomial, P5(x), at x0 = 0 for cosx. Determine an upper bound for the error |P5(x) − cosx| for x ∈ [−π/2,π/2]. 4.5.3 Use Theorem 4.5.2 to determine if the following functions have a local minimum or a local maximum at the indicated points. (a) f (x)= x3 sinx at x 0 = 0. (b) f (x)=(1 − x)lnx at x0 = 1. 4.5.4 Suppose f is twice differentiable on (a,b). Show that for every x ∈ (a,b), f (x + h)+ f (x− h) − 2 f (x) ′′ (x). lim = f h2 h →0 4.5.5 ▶ (a) Suppose f is three times differentiable on (a,b) and x0 ∈ (a,b). Prove that ′ (x0) h ′′ (x 0) h2 ′′′ (x0) f (x0 + h) − f (x0) − f f 1! − f 2! lim = . h3 h →0 3! (b) State and prove a more general result for the case where f is n times differentiable on (a,b). 4.5.6 Suppose f is n times differentiable on (a,b) and x0 ∈ (a,b). Defne n hn Pn(h)= f (n)(x 0) for h ∈ R. ∑ n! k=0
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