171 It follows that f is differentiable on R with 2 − 1 x3 e x2 , if x ≠ 0; f ′ (x)= 0, if x = 0. In a similar way, we can show that f is twice differentiable on R with 6 2 − 1 e x , if x ≠ 0; ′′ (x)= x4 + x6 2 f 0, if x = 0. Based on these calculations, we predict that f is n times differentiable for every n ∈ N with 1 − 1 2 P e x , if x ≠ 0; f (n)(x)= x 0, if x = 0, where P is a polynomial. Now we proceed to prove this conclusion by induction. The conclusion is true for n = 1 as shown above. Given that the conclusion is true for some n ∈ N, for x ≠ 0 we have 1 2 1 − 1 1 − 1 2 2 f (n+1)(x)= −x−2P ′ + P e x = Q e x , x3 x x x where Q is also a polynomial. It is an easy exercise to write the explicit formula of Q based on P. Moreover, successive applications of l’Hôpital’s rule give f (n)(x) − f (n)(0) 1 1 1 tP(t) 2 lim = lim P e − x = lim = 0. x→0+ x − 0 x→0+ x x t→∞ et 2 In a similar way, we can show that f (n)(x) − f (n)(0) lim = 0. x→0− x − 0 Therefore, f (n+1)(0)= 0. We have proved that for every n ∈ N, f is n times differentiable and, so, f ∈ Cn(R). Here we do not need to prove the continuity of f (n) because the differentiability of f (n) implies its continuity. In a similar way, we can also show that the function ( e− 1 x , if x > 0; f (x)= 0, if x ≤ 0 is n times differentiable for every n ∈ N. SECTION 4.5 Exercise 4.5.1. Let f (x)= ex . By Taylor’s theorem, for any x > 0, there exists c ∈ (0,x) such that m f (k)(0) f (m+1)(c) x k m+1 f (x)= e = ∑ x + c k=0 k! (m+ 1)! m k m k x x ∑ e c m+1 ∑ = + c > . k=0 k! (m + 1)! k! k=0
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