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Taylor's Theorem

  1. Jan 16, 2015 #1
    I am studying power series right now and I am understanding well how to write them and where they converge but I am having some trouble grasping the Taylor Remainder Theorem for a few reasons.

    First of all it says the remainder is:

    f^(n+1)(c)(x-a)^(n+1)/(n+1)! for some c between a and x.

    I do not know what c it means, how do I know what value to use as c?

    My next issue is that for power series for sinx and cosx:

    For example if I am doing sinx and use three terms x-x^3/3!+x^5/5! My previous derivative value that goes in front of the x^5/5! is 1 because cos(0)=1 but taylor's theorem states f^(n+1) so the next derivative would be -sin(0)=0. So would that mean the remainder theorem would be 0 in this case?

    Something else I am confused about is what my book calls the "remainder estimation theorem" where it says if absolute value of f^(n+1)(t) is equal to or less than Mr^(n+1) then Rnx is = or < Mr^(n+1)(x-a)^(n+1)/(n+1)!
    My book does 2 examples but it does not explain very well how to choose these values of M and r and how exactly I can use it to estimate.

    I have read this chapter twice and either because of the way they word it or because I am thinking about it wrong I am unable to understand this remainder theorem which is frustrating because I feel good on the rest of the power series topic except for this. Thank you for any help.
  2. jcsd
  3. Jan 16, 2015 #2


    Staff: Mentor

    The remainder theorem doesn't specify an exact value - it merely says that c is a number between a and x. You want the remainder expression to be as large as you make it, so as to be able to specify the largest your error could be. One way to do this is to look at f(n + 1)(x). If it happens to be an increasing function, the largest value of this function will come at the right endpoint of the interval. If the function happens to be decreasing, the largest value occurs at the left endpoint of the interval. For other functions, you try to estimate the largest value in the interval.
    No. As before, the remainder is f(n + 1)(c) * xn + 1/(n + 1)!. IOW, you don't evaluate this derivative at 0, but at some point between 0 and x.
    It would be helpful for you to show us one of these examples, as well as less work to explain.
  4. Jan 16, 2015 #3
    For an example of the M and r:

    The example says estimate the maximum value of the error for x-x^2/2 which is used to estimate ln(1+x) on the interval -0.1 to 0.1 inclusive. It starts by saying the maximum value of the third derivative ( 2/(1+t)^3) occurs at -0.1 and is equal to 2000/729, this is the value they give M. Then they say "we can let r=1" and I am not sure why we do. Then they say the remainder is equal to or less than (2000/729)(absolute value x)^3 /3! Then they plug in + or - 0.1 into x and end up with remainder being approx. 4.6x10^-4. I understand that M is the max value of the third derivative over the interval but I do not know what "r" is in the first place and therefore do not understand why the say they can let it equal 1, and also I do not understand why we plug in + or - 0.1 into x? Do we plug it in because it is the bounds? because that doesn't make sense why we would plug the bounds into x for the remainder to me.
  5. Jan 16, 2015 #4


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    It's not a question of evaluating the remainder as much as bounding the remainder. The normal use of the remainder formula is to bound it so you know that you have expanded enough terms and can ignore the remainder, whatever it is. For your earlier example of sin(x), you know all derivatives and can easily bound the remainder.
  6. Jan 16, 2015 #5
    A few more questions:

    When we say that we are trying to find a maximum value for f^(n+1)(c) does that mean maximum absolute value or highest positive? Can the remainder be negative?

    Also, when trying to use this formula I am still unsure what value to plug in for x. I have learned that for f^(n+1)(c) the c we use must be where the maximum occurs but what value do we plug in for the x in the (x-a)^n+1 part of the theorem?

    Also, I came up with a problem that sort of demonstrates why I am having trouble:
    If I want to find the remainder of evaluating sin3 by using -(x-pi) + (x-pi)^3/3! (The 2nd order series for sinx center at pi) I am not sure what I would use.

    F^(n+1)(c) would be sin(c) but I'm not exactly sure where to find maximum value because there is not really an interval

    the rest would be (x-pi)^4/4! but I do not know what value to plug into x for the remainder?
  7. Jan 16, 2015 #6


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    The next part of the expansion would be (x-pi)^5/5! . Therefore your error estimate would be abs(a-pi)^5/5! where a is an unspecified value somewhere between 0 and x. In this case, the largest value for abs(a-pi) with a between 0 and x is when a = x.
  8. Jan 16, 2015 #7


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    You are given a value of x that you want to use a partial Taylor series to determine f(x) within a certain accuracy. If you need a certain amount of accuracy from a partial expansion of the Taylor series, you need to expand enough Taylor series terms to guarantee the accuracy you need. How many terms do you need? The bounds on the remainder tell you. You are only bounding the remainder to the accuracy needed. You are not finding the remainder's value.
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