Question

use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange remainder

Answer 1

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: