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Lagrange remainder theorem

  1. Feb 3, 2015 #1
    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
  2. jcsd
  3. Feb 3, 2015 #2


    Staff: Mentor

    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.
  4. Feb 4, 2015 #3
    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.
  5. Feb 4, 2015 #4


    Staff: Mentor

    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.
  6. Feb 4, 2015 #5
    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.
  7. Feb 4, 2015 #6


    Staff: Mentor

    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.
  8. Feb 4, 2015 #7
    Oh ok I see what you mean thank you for clarifying.
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