How can the Lagrange remainder theorem be applied to series with skipped terms?

However, the exact values of c and x will depend on the specific function and interval you are working with. It may be helpful to try out different values and see what gives you the smallest error.
  • #1
member 508213
I have a few questions about the remainder theorem.

1: For series that "skip" terms (example: 1+x^2+x^4+x^6) the theorem says the n+1 derivative and x^(n+1)/(n+1)!. For example if you have 1 + x^2 where you know the next term would be x^4 you could treat it as a third order or a second order. My book has situations where they do both variations and it is extremely confusing because I do not know which to choose.

Example: This is a problem I made up: Estimate error involved in using 1+x^2 + (x^4)/2 to estimate e^(x^2) on -0.1 to 0.1.

So do I treat it as forth order or fifth order? Because my book has situations similar to this where it has done both so I have no idea what to do? It asks for n+1 derivative so is it the fifth or six derivative? Do I do (.1)^6/6! or (.1)^5/5!.

This is a major issue I am having please explain why it is one or the other because I can see why it could go either way.

2: The best way to ask my second question is to just show another example it is difficult to explain.

Example: Use x-(x^2)/2 to estimate ln(x+1) over the interval [0, 0.2].

The theorem states f^(n+1)(c) (x)^(n+1) / (n+1)!

Here is what I believe: The c value and the x value can be different in order to maximize the different parts. Is this true?

For this example I would have c be 0 in order to maximize the third derivative of ln(x+1) which is 2/(x+1)^3...but I would have x be .2 in order to maximize the (x)^3 part of the theorem?

Is what I assumed correct? You can have different values of c and x in order to maximize the different parts? I believe this is correct so hopefully you can confirm this, if not can you please explain why? Thank you very much for replies, both of these questions have been confusing to me and any help is appreciated
 
Physics news on Phys.org
  • #2
Austin said:
I have a few questions about the remainder theorem.

1: For series that "skip" terms (example: 1+x^2+x^4+x^6) the theorem says the n+1 derivative and x^(n+1)/(n+1)!. For example if you have 1 + x^2 where you know the next term would be x^4 you could treat it as a third order or a second order. My book has situations where they do both variations and it is extremely confusing because I do not know which to choose.
The higher the degree is in the last term of your approximating polynomial, the smaller the error will be, so the short answer is use the higher order.
Austin said:
Example: This is a problem I made up: Estimate error involved in using 1+x^2 + (x^4)/2 to estimate e^(x^2) on -0.1 to 0.1.

So do I treat it as forth order or fifth order? Because my book has situations similar to this where it has done both so I have no idea what to do? It asks for n+1 derivative so is it the fifth or six derivative? Do I do (.1)^6/6! or (.1)^5/5!.

This is a major issue I am having please explain why it is one or the other because I can see why it could go either way.

2: The best way to ask my second question is to just show another example it is difficult to explain.

Example: Use x-(x^2)/2 to estimate ln(x+1) over the interval [0, 0.2].

The theorem states f^(n+1)(c) (x)^(n+1) / (n+1)!
That's not what it states. It says something about the expression above; namely, that it is the error in approximating the function with terms up to degree n.
Austin said:
Here is what I believe: The c value and the x value can be different in order to maximize the different parts. Is this true?
Sure -- there is nothing that says that they have to be equal. The theorem in question is a sort of existence theorem -- it says that "there exists a number c" in a given interval such that if you truncate the Taylor series at the term of degree n, then the exact value of the remainder is such and such. It DOES NOT tell you what that magic number c is, though, but if you have a small enough interval, you can get an idea of how close you are.

For a given function f, if f(n + 1) happens to be either increasing or decreasing, you can easily tell where the maximum value of f(n + 1) will be -- at one or the other endpoint. For other functions, you might have to do something more creative to figure out where the maximum value is.
Austin said:
For this example I would have c be 0 in order to maximize the third derivative of ln(x+1) which is 2/(x+1)^3...but I would have x be .2 in order to maximize the (x)^3 part of the theorem?

Is what I assumed correct? You can have different values of c and x in order to maximize the different parts? I believe this is correct so hopefully you can confirm this, if not can you please explain why?Thank you very much for replies, both of these questions have been confusing to me and any help is appreciated
 
  • #3
Mark44 said:
The higher the degree is in the last term of your approximating polynomial, the smaller the error will be, so the short answer is use the higher order.
That's not what it states. It says something about the expression above; namely, that it is the error in approximating the function with terms up to degree n.
Sure -- there is nothing that says that they have to be equal. The theorem in question is a sort of existence theorem -- it says that "there exists a number c" in a given interval such that if you truncate the Taylor series at the term of degree n, then the exact value of the remainder is such and such. It DOES NOT tell you what that magic number c is, though, but if you have a small enough interval, you can get an idea of how close you are.

For a given function f, if f(n + 1) happens to be either increasing or decreasing, you can easily tell where the maximum value of f(n + 1) will be -- at one or the other endpoint. For other functions, you might have to do something more creative to figure out where the maximum value is.
Thank you for your reply it was helpful. However, can you explain what you meant by "That's not what it states. It says something about the expression above; namely, that it is the error in approximating the function with terms up to degree n." This part confused me and I just wanted to understand what you were saying here.
 
  • #4
Austin said:
Thank you for your reply it was helpful. However, can you explain what you meant by "That's not what it states. It says something about the expression above; namely, that it is the error in approximating the function with terms up to degree n." This part confused me and I just wanted to understand what you were saying here.

You said this:
Austin said:
The theorem states f^(n+1)(c) (x)^(n+1) / (n+1)!
The theorem states that the above is the remainder, or error, if you truncate the infinite Taylor series at the term of degree n.
 
  • #5
Mark44 said:
You said this:
The theorem states that the above is the remainder, or error, if you truncate the infinite Taylor series at the term of degree n.

I'm sorry I think I'm missing what you are saying here... when I said "the theorem states" I typed the formula for the remainder after that. What did I say wrong? Sorry I am just trying to fully understand what you are saying.
 
  • #6
You were unclear in what you said, saying that
The theorem states f^(n+1)(c) (x)^(n+1) / (n+1)!
You didn't say that this was the remainder or any of the other stuff that the theorem actually states. That was my point.
 
  • #7
Mark44 said:
You were unclear in what you said, saying that You didn't say that this was the remainder or any of the other stuff that the theorem actually states. That was my point.
Oh ok I see what you mean thank you for clarifying.
 

What is the Lagrange remainder theorem?

The Lagrange remainder theorem, also known as the Taylor remainder theorem, is a mathematical tool used to estimate the error in approximating a function with a polynomial of a certain degree. It is named after the mathematician Joseph-Louis Lagrange.

How is the Lagrange remainder theorem used?

The Lagrange remainder theorem is used to determine the accuracy of polynomial approximations of functions. It provides an upper bound for the difference between the actual value of a function and its approximation at a given point. This can be useful in many areas of mathematics and engineering, such as numerical analysis and optimization.

What is the formula for the Lagrange remainder theorem?

The formula for the Lagrange remainder theorem is:
R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}
where f(x) is the function being approximated, a is the center of the approximation, n is the degree of the polynomial, and c is a value between a and x.

What are the applications of the Lagrange remainder theorem?

The Lagrange remainder theorem has many applications in mathematics and engineering. It is used in numerical analysis to estimate the error in approximating a function, in optimization to find the maximum or minimum of a function, and in physics to model complex systems using polynomial approximations.

How does the Lagrange remainder theorem relate to the Taylor series?

The Lagrange remainder theorem is a special case of the Taylor series. It provides an estimate for the remainder term in the Taylor series, which is the difference between the actual value of a function and its approximation. The Taylor series is a powerful tool for approximating functions, and the Lagrange remainder theorem helps to determine the accuracy of these approximations.

Similar threads

Replies
3
Views
1K
Replies
11
Views
2K
Replies
14
Views
2K
  • Calculus
Replies
12
Views
2K
  • Precalculus Mathematics Homework Help
Replies
3
Views
848
Replies
1
Views
936
  • Calculus
Replies
5
Views
957
Back
Top