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 n be a positive integer. Suppose f : [a,b] → R is 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 any x ∈ [a,b] with x ̸ = x0, there exists a number c in between x0 and x such that f (n+1)(c) (x− x0) n+1 f (x)= Pn(x)+ , (n+ 1)! where n f (k)(x 0) Pn(x)= (x − x0) k . k ∑ =0 k! Proof: Let x0 be as in the statement and fx x ≠ x0. Since x − x0 ≠ 0, there exists a number λ ∈ R such that λ (x − x0) n+1 f (x)= Pn(x)+ . (n + 1)! We will now show that λ = f (n+1)(c), for some c in between x 0 and x. Consider the function n f (k)(t) λ g(t)= f (x) − ∑ (x −t)k − (x −t)n+1 . k=0 k! (n + 1)! Then n f (k)(x 0) (x−x0) k − λ (x−x0) n+1 λ (x−x0) n+1 g(x 0)= f (x)− ∑ = f (x)−Pn(x)− = 0. k=0 k! (n + 1)! (n + 1)! and n f (k)(x) λ g(x)= f (x) − ∑ (x − x)k − (x − x)n+1 = f (x) − f (x)= 0. k=0 k! (n + 1)! By Rolle’s theorem (Theorem 4.2.2), there exists c in between x0 and x such that g ′ (c)= 0. Taking the derivative of g (keeping in mind that x is fxed and the independent variable is t) and using the product rule for derivatives, we have ! n f (k+1)(c) (x− c)k f (k)(c) λ g ′ (c) = − f ′ (c)+ ∑ − + (x− c)k−1 + (x − c)n k=1 k! (k − 1)! n! = λ (x − c)n − 1 f (n+1)(c)(x − c)n = 0. n! n! 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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