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Error estimate for Taylor polynomials

  1. Feb 28, 2012 #1
    Use the error estimate for Taylor polynomials to find an n such that

    | e - (1 + (1/1!) + (1/2!) + (1/3!) + ... + (1/n!) | < 0.000005

    all i have right now is the individual components...

    f(x) = ex
    Tn (x) = 1/ (n-1)!

    k/(n-1)! |x-a|n+1 = 0.000005
    a = 0
    x = 1

    I don't know where to go from here
  2. jcsd
  3. Feb 28, 2012 #2


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    The max error if your last term is$$
    \frac {f^{(n)}(a)(x-a)^n}{n!}$$ is$$
    \left | \frac {f^{(n+1)}(c)(x-a)^{n+1}}{(n+1)!}\right |$$As you have already observed ##a=0## and ##x = 1##. How big can that derivative term be for ##0\le c\le 1##? Once you have that, figure out how large ##n## needs to be to make it small enough.
  4. Feb 28, 2012 #3
    how do i figure out c?
  5. Feb 28, 2012 #4
    i just looked over my notes again. we started this problem in class, and our professor told us to pick a value for k above the value of e.

    so if k is 3,
    3/(n+1)! = 0.000005

    so now i have to find n?
  6. Feb 29, 2012 #5


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    You don't have to "figure out c". You have the nth derivative of your function evaluated at c, and c is in the interval [0,1]. You don't know the exact value of c so you ask yourself, "how big can ##f^{(n)}(c)= e^c## be for c in [0,1]. Do you understand why your instructor says to pick k > e?

    To answer your last question, you don't want 3/(n+1)! = 0.000005. You would be very lucky to find an integer n giving equailty. You want 3/(n+1)! < 0.000005. Writing it a different way, you want$$
    (n+1)! > \frac 3 {.000005}$$Factorials grow very quickly. Is shouldn't be difficult to check by hand how big n needs to be.
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