# Lagrange remainder

## Homework Statement

This is not a homework problem but I would like to clarify my concern.

It is stated that a function can be written as such:

$f(x) = \lim_{n \rightarrow ∞} \sum^{∞}_{k=0} f^{(k)} \frac{(x-x_{0})^k}{k!}$

$R_{n}=\int^{x}_{x_{0}} f^{(n+1)} (t) \frac{(x-t)^n}{n!} dt$

They state that by MVT,

$R_{n}= \frac{f^{(n+1)}(x^{*})}{(n+1)!} (x-x_{0})^{n+1}$

For some $x^{*} \in (x_{0},x)$

I am wondering which statement of MVT leads to the second identity? Much thanks:)

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
If MVT means "mean value theorem", then there is one about ## \int_a^b f(x)g(x) dx ##.

If MVT means "mean value theorem", then there is one about ## \int_a^b f(x)g(x) dx ##.
Yes, I am aware that MVT is mean value theorem and it is related to the integral form of MVT. But I am wondering how did they manage to get the following result.

The first mean value theorem states:

$\int^{b}_{a} G(t) dt = G(x) (b-a)$ for $x \in (a,b)$

Which does not lead to the result desired.

Therefore, I am convinced that it is related to the second mean value theorem:

If G : [a, b] → R is a monotonic (not necessarily decreasing and positive) function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b) such that

$\int^{b}_{a} G(t)\phi(t) dt = G(a+)\int^{x}_{a}\phi(t) dt + G(b-)\int^{b}_{x}\phi(t) dt$

However, I am not sure how do I go about deal with that

Yes, i have took the theorem stated above form the similar website

So you are saying you cannot see a useful one there?

I am convinced that it is related to the second mean value theorem:

If G : [a, b] → R is a monotonic (not necessarily decreasing and positive) function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b) such that

∫baG(t)ϕ(t)dt=G(a+)∫xaϕ(t)dt+G(b−)∫bxϕ(t)dt

However, I am not sure how do I go about deal with that

I fail to see why what is labeled as the "First mean value theorem for integration" in the article I linked is not suitable.

You are free to make it much more complex than it needs to be, of course.