111 4.5 Taylor’s Theorem In this section, we prove a result that lets us approximate differentiable functions by polynomials. Theorem 4.5.1 — Taylor’s Theorem. Let nbe a positive integer. Suppose f : [a,b] →Ris a function such that f (n) is continuous on [a,b], and f (n+1)(x) exists for all x ∈(a,b). Let x 0 ∈[a,b]. Then for anyx ∈[a,b] withx̸ =x0, there exists a number c in betweenx0 andx such that f (x) =Pn(x)+ f (n+1)(c) (n+1)! (x−x0) n+1, where Pn(x) = n ∑ k=0 f (k)(x 0) k! (x−x0) k. Proof: Let x0 be as in the statement and fixx̸ =x0. Since x−x0̸ =0, there exists a number λ ∈R such that f (x) =Pn(x)+ λ (n+1)! (x−x0) n+1. We will now show that λ =f (n+1)(c), for some c in betweenx 0 andx. Consider the function g(t) =f (x)− n ∑ k=0 f (k)(t) k! (x−t)k − λ (n+1)! (x−t)n+1. Then g(x0) =f (x)− n ∑ k=0 f (k)(x 0) k! (x−x0) k− λ (n+1)! (x−x0) n+1 =f (x)−Pn(x)− λ (n+1)! (x−x0) n+1 =0. and g(x) =f (x)− n ∑ k=0 f (k)(x) k! (x−x)k − λ (n+1)! (x−x)n+1 =f (x)−f (x) =0. By Rolle’s theorem (Theorem 4.2.2), there exists c in betweenx0 andx such that g′(c) =0. Taking the derivative of g(keeping in mind that x is fixed and the independent variable is t) and using the product rule for derivatives, we have g′(c) = −f ′(c)+ n ∑ k=1 − f (k+1)(c) k! (x−c)k + f (k)(c) (k−1)! (x−c)k−1!+ λ n! (x−c)n = λ n! (x−c)n − 1 n! f (n+1)(c)(x−c)n =0. It follows that λ =f (n+1)(c). The proof is now complete. □ The polynomial Pn(x) given in the theorem is called the nth Taylor polynomial of f at x0.
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