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Lagrange remainder

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data

    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:)
     
    Last edited: Apr 14, 2013
  2. jcsd
  3. Apr 14, 2013 #2
    If MVT means "mean value theorem", then there is one about ## \int_a^b f(x)g(x) dx ##.
     
  4. Apr 14, 2013 #3
    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. Apr 14, 2013 #4
  6. Apr 14, 2013 #5
    Yes, i have took the theorem stated above form the similar website
     
  7. Apr 14, 2013 #6
    So you are saying you cannot see a useful one there?
     
  8. Apr 14, 2013 #7
    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
     
  9. Apr 14, 2013 #8
    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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