Lagrange remainder

  • Thread starter icystrike
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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:

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

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

They state that by MVT,

[itex] R_{n}= \frac{f^{(n+1)}(x^{*})}{(n+1)!} (x-x_{0})^{n+1} [/itex]

For some [itex] x^{*} \in (x_{0},x) [/itex]

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

  • #2
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If MVT means "mean value theorem", then there is one about ## \int_a^b f(x)g(x) dx ##.
 
  • #3
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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:

[itex] \int^{b}_{a} G(t) dt = G(x) (b-a) [/itex] for [itex] x \in (a,b) [/itex]

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

[itex] \int^{b}_{a} G(t)\phi(t) dt = G(a+)\int^{x}_{a}\phi(t) dt + G(b-)\int^{b}_{x}\phi(t) dt [/itex]

However, I am not sure how do I go about deal with that
 
  • #5
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Yes, i have took the theorem stated above form the similar website
 
  • #6
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So you are saying you cannot see a useful one there?
 
  • #7
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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
 
  • #8
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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.
 

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