24.4k views
4 votes
use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange remainder

User Al Dass
by
8.3k points

1 Answer

3 votes

Answer:

As consequence of the Taylor theorem with integral remainder we have that


f(x) = f(a) + f'(a)(x-a) + (f''(a))/(2!)(x-a)^2 + \cdots + (f^((n))(a))/(n!)(x-a)^n + \int^a_x f^((n+1))(t)((x-t)^n)/(n!)dt

If we ask that
f has continuous
(n+1)th derivative we can apply the mean value theorem for integrals. Then, there exists
c between
a and
x such that


\int^a_x f^((n+1))(t)((x-t)^k)/(n!)dt = (f^((n+1))(c))/(n!) \int^a_x (x-t)^n d t = (f^((n+1))(c))/(n!) ((x-t)^(n+1))/(n+1)\Big|_a^x

Hence,


\int^a_x f^((n+1))(t)((x-t)^k)/(n!)d t = (f^((n+1))(c))/(n!) ((x-t)^((n+1)))/(n+1) = (f^((n+1))(c))/((n+1)!)(x-a)^(n+1) .

Thus,


\int^a_x f^((n+1))(t)((x-t)^k)/(n!)d t = (f^((n+1))(c))/((n+1)!)(x-a)^(n+1)

and the Taylor theorem with Lagrange remainder is


f(x) = f(a) + f'(a)(x-a) + (f''(a))/(2!)(x-a)^2 + \cdots + (f^((n))(a))/(n!)(x-a)^n + (f^((n+1))(c))/((n+1)!)(x-a)^(n+1).

Explanation:

User Zergatul
by
7.5k points