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I Regarding Error Bound of Taylor Series

  1. May 17, 2016 #1
    Hi all, I am very confused about how one can find the upper bound for a Taylor series.. I know its general expression, which always tells me to find the (n+1)th derivative of a certain function and use the equation f(n+1)(c) (x-a)n+1/(n+1)! for c belongs to [a,x]

    However, there are numerous cases in which I am only given a sigma summation form of an unknown function. If I want to find the nth degree error term (assume nth degree is the nth term in the summation), do I simply take the (n+1)th term? I personally do not agree with this since the n+1 th term is not necessarily the biggest. In normal situations I always re-evaluate the n+1 th derivative and find its max value, but with unknown functions I am not able to do that.

    Any explanation on how to find the error bound? Do I need to find the n+1 th derivative every time?
     
  2. jcsd
  3. May 18, 2016 #2

    Mark44

    Staff: Mentor

    Can you give one example of the kind you're talking about? It seems to me that you might be talking about a problem where you are given a series, and are asked to find an upper bound for the remainder if you use the first n terms in the series.
    Yes, if you are working with a Taylor series, but all you need is the largest value of this derivative on the interval [a, x], as you said. An example would be very helpful for me to understand what you're asking.
     
  4. May 18, 2016 #3
    A simple example would be maclaurin expansion of f(x)=sin(x) given the interval [0, 0.5]. Assume I don't know that this is the expansion for sin(x), all I know is the series. Now I am asked to write down the expansion to degree 3, and find the approximate value of this function at f(0.5). Looking for the approximate value part is obvious to me, but when it asks me to find the upper bound to its error, I can't just take the next term since in the given interval the 4th derivative of sin(x) will not reach 1. So in this case I need to find the 4th derivative of the original function. But as I assumed, if I do not know the original function but only know its maclaurin form, how can I find the lagrange error?
     
  5. May 18, 2016 #4

    Mark44

    Staff: Mentor

    Which is an alternating series, right? There's a theorem that you can use to find the remainder when you use the first n terms of an alternating series.
     
  6. May 18, 2016 #5
    I don't think it's an alternating series.. I know what you are talking about though. I think you are talking about truncation error where I can just take the next term as the error bound for an alternating series. But I am talking about Lagrange error of nth degree expansion.. I don't know if this is clear.
     
  7. May 18, 2016 #6

    Mark44

    Staff: Mentor

    The Maclaurin series for both sin(x) (=## x - \frac{x^3}{3!} + \frac{x^5}{5!} -+ \dots##) and cos(x) (=## 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -+ \dots##)are alternating series.

    Regarding your question, if you're expected to find the Lagrange error term, you would need to know the function whose derivatives appear in the Taylor expansion. Without the underlying function, I don't see how you would be able to get ##f^{(n)}(c)##.
     
  8. May 18, 2016 #7
    So Lagrange error always require me to find the derivative. I can't just take the next term (if the maclaurin or taylor expansion is not alternating) as my answer?
     
  9. May 18, 2016 #8

    Mark44

    Staff: Mentor

    Right on both counts.
     
  10. May 21, 2016 #9
    Ok! Thanks a lot!
     
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