1549901369-Elements_of_Real_Analysis__Denlinger_

6.5 Taylor's Theorem 335

Proof. Suppose f is n times differentiable on an open interval containing a

and x, where x -:j:. a, and J(n+l) (t) exists for all tin the open interval I between

a and x. Suppose Tn(x) and Rn(x) are as defined above and define the function

G ( t) on I by the formula

n J(k)(t) k (x -tr+l

G(t) = f(t) + L ~(x -t) + Rn(x) (x _ a)n+l · (6)

k=l

Then G is differentiable at every t E I and continuous on the closure of
I , the closed interval between a and x. Remembering that x is constant in
Equation (6), the derivative of G is (fill in reasons for each step):

G'(t) = J'(t) + t [JC:?) k(x - t)k-^1 (-1) + JCk::l(t) (x -t)k]

k=l

(n+l)(x-t)n(-l) R ()

+ (x - a)n+l n x

n J(k)( ) n J(k+l)( )

= J'(t) - "°""" t (x - t)k-l + "°""" t (x - t)k

L., (k - 1)! L., k!

k=l k=l

_ (n + l)(x - t)n Rn(x)

(x -a)n+l

n-l J(k+l)( ) n J(k+l)( )

= f'(t) - "°""" t (x - t)k + "°""" t (x -t)k

L., k! L., kl

k=O k=l

_(n+l)(x-t)nR (x)

(x - a)n+l n

= J'(t) - f'(t) (x -t)o + J(n+l)(t) (x - t)n - (n + l)(x -t)n Rn(x).

O! n! (x - a)n+l

1 f(n+l)(t) n (n + l)(x -t)n
Thus, G (t) = n! (x -t) - (x _ a)n+l Rn(x). (7)

(Remember, in Equation (6), tis the independent variable and xis constant.)

Now, letting t = a in Equation (6), we find

n J(k)(a) (x -a)n+l

G(a) = f(a) + L ~(x -a)k + Rn(x) (x _ a)n+l = Tn(x) + Rn(x)

k=l

= f(x), by Definition 6.5.9.