4. Taylor’s Theorem for Functions of One Variable: Suppose that f ∈ Cn[a, b]andthe (n+1) derivative f exists on [a, b], and let x0 ∈ [a, b]. For every x ∈ [a, b] there exists η(x) between x0 and x such that
(2) (n) (n+1) (1) f (x0) 2 f (x0) n f (η) n+1 f(x)=f(x0)+f (x0)(x−x0)+ (x−x0) +···+ (x−x0) + (x−x0) . 2! n! (n +1)!
Proof Use integration by parts: x (1) f(x)=f(x0)+ f (t) dt x0 x (1) x (2) = f(x0) − f (t)(x − t) + f (t)(x − t) dt x 0 x0 x (1) (2) 1 2 (n) 1 n = f(x0) − f (t)(x − t)+f (t) (x − t) + ...+ f (t) (x − t) 2 n! x0 1 x + f (n+1)(t)(x − t)n dt n! x0 (1) (2) 1 2 (n) 1 n = f(x0)+f (x0)(x − x0)+f (x0) (x − x0) + ...+ f (x0) (x − x0) 2 n! 1 x + f (n+1)(t)(x − t)n dt . n! x0 n The remainder term is in Young’s integral form. Noting that (x − t) has one sign on [x0,x], application of the generalised integral mean value theorem gives