Lagrange Remainder: Clarifying MVT Statement

In summary, the conversation revolved around the use of the mean value theorem in relation to the integral form. The person was wondering which statement of the mean value theorem leads to the second identity. After some discussion and clarification, it was determined that the second mean value theorem is the relevant one. However, it was not clear how to apply this theorem to obtain the desired result. Ultimately, it was suggested to use the first mean value theorem for integration, which states that for a monotonic function and an integrable function, there exists a number within the interval where the integral can be expressed as a combination of the integrands at the endpoints.
  • #1
icystrike
445
1

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:)
 
Last edited:
Physics news on Phys.org
  • #2
If MVT means "mean value theorem", then there is one about ## \int_a^b f(x)g(x) dx ##.
 
  • #3
voko said:
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
Yes, i have took the theorem stated above form the similar website
 
  • #6
So you are saying you cannot see a useful one there?
 
  • #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
 
  • #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.
 

Related to Lagrange Remainder: Clarifying MVT Statement

1. What is the Lagrange remainder in the context of the Mean Value Theorem (MVT)?

The Lagrange remainder is a term used in the proof of the Mean Value Theorem to represent the difference between the actual value of a function at a given point and the value predicted by the MVT. It is often denoted by the letter R and is equal to the difference between the function and its tangent line at a specific point.

2. How is the Lagrange remainder calculated?

The Lagrange remainder is calculated using the formula R = f'(c)(x-a), where f'(c) is the derivative of the function at some point c between a and x, and (x-a) is the distance between x and a. This formula is derived from the MVT and is used to determine the error or deviation from the predicted value of the function at a specific point.

3. What is the significance of the Lagrange remainder in the MVT?

The Lagrange remainder is significant in the MVT as it provides a way to quantify the error or deviation from the predicted value of a function at a specific point. It allows us to determine the maximum possible error in using the MVT to approximate a function and is therefore important in understanding the limitations of the MVT.

4. Can the Lagrange remainder be negative?

Yes, the Lagrange remainder can be negative. This means that the actual value of the function at a given point is less than the predicted value by the MVT. This can happen when the function is concave down and the tangent line overestimates the function's value, resulting in a negative remainder.

5. How is the Lagrange remainder used in practice?

The Lagrange remainder is primarily used in theoretical mathematics to prove the MVT and other related theorems. In practical applications, it is rarely used directly but is important in understanding the accuracy and limitations of using the MVT to approximate a function. It is also used in numerical analysis to estimate the error in numerical methods for finding solutions to equations.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
639
  • Calculus and Beyond Homework Help
Replies
1
Views
489
  • Calculus and Beyond Homework Help
Replies
3
Views
478
  • Calculus and Beyond Homework Help
Replies
3
Views
613
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
582
  • Calculus and Beyond Homework Help
Replies
6
Views
817
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
626
Back
Top