Lagrange Remainder: Clarifying MVT Statement

[icystrike]

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:)

 

[voko]

If MVT means "mean value theorem", then there is one about $\int_a^b f(x)g(x) dx$.

 

[icystrike]

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

 

[voko]

I think you are mistaken about "first" and "second". Anyway, have a look here: http://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_integration

Can you spot a useful one?

 

[icystrike]

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

 

[voko]

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

 

[icystrike]

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

 

[voko]

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.