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Taylor polynomial remainder term

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the followign function [itex] f(x) = x^-5 [/itex]
    [itex] a=1 [/itex]
    [itex] n=2 [/itex]
    [itex]0.8 \leq x \leq 1.2 [/itex]

    a) Approximate f with a tayloy polynomial of nth degree at the number a = 1
    b) use taylor's inequality to estimate the accuracy of approximation [itex] f(x) ≈ T_{n}(x) [/itex] when x lies in the interval

    2. Relevant equations
    [tex] f(x) = f(a) + f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2 [/tex]
    [tex] R_k(x) =\frac{f^{(k+1)}(\xi_L)}{(k+1)!} (x-a)^{k+1} [/tex]

    3. The attempt at a solution

    Part a is simple:
    [tex] T_2(x) = 1 - 5(x-1) + 15(x-1)^2 [/tex]

    Since we have found the taylor polynomial at n = 2 the remainder:

    [tex] R_2(x) \leq | \frac{M}{3!}(x-1)^3 | [/tex]

    Since [tex] f^{(3)} (x) = -210x^{-8} [/tex]
    and this is decreasing, we use x = 0.8 and we use [itex] M = -210(0.8)^{-8} [/itex]

    [tex] R_2(x) \leq |\frac{-210(0.8)^{-8}}{3!} (0.8-1)^3 [/tex]

    and the result of the above is 1.6689

    Is the above correct? Thanks for your help!
     
  2. jcsd
  3. Feb 28, 2013 #2

    eumyang

    User Avatar
    Homework Helper

    Technially,
    [itex] f^{(3)} (x) = -210x^{-8} [/itex]
    is increasing on the interval. But since we need
    [itex]M = max \left| f^{(k+1)}(\xi_L) \right|[/itex]
    x = 0.8 is the correct value of xi.

    I also got 1.6689 as the estimated error.
     
  4. Feb 28, 2013 #3
    Ah yes, didn't consider the negative sign - thanks for your help!
     
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