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Taylor formula

  1. Nov 26, 2004 #1
    1. Let f(x) = (1+x)^{2/3}

    (a) find the taylor polynomial T_2(x) of f expanded about a = 0.

    i got 1 + (1/3)x - (1/9)x^{2}

    For the rest, i have no idea how to do...any help would be greatly appreciated.

    (b) For the givven f write the lagrange remainder formula for the error term f(x) - T_2(x).

    (c) Show that when x>0 the error f(x)-T_2(x) is at most (5/18)x^{3}. and

    (d) Write a fraction that estimates (1.2)^{1/3}, and show that the error in your estimate is at most 1/2025.

    Thanks for any help you can provide!
  2. jcsd
  3. Nov 27, 2004 #2


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    What did you get for f'(x) and f''(x)? Your polynomial is a bit off.

    Do you know what the Lagrange remainder is? If not look it up and post again with what about it is troubling you.
  4. Nov 27, 2004 #3
    sorry, T_2(x) is suppose to be 1 + (2/3)x - (1/9)x^{2}

    i believe the lagrange remainder is: f^{n+1}(p) * (x-a)^{n+1}

    where p is between (a,n).
    The problem is, i don't know how to find the last term in the sequence or the error (which i believe is also called the lagrange remainder, right?) thanks for your reply!
  5. Nov 27, 2004 #4
    sorry, the lagrange remainder is suppose to be

    f^{n+1}(p) * (x-a)^{n+1}
  6. Nov 27, 2004 #5


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    The remainder term you have gives f(x)-T_n(x), the difference between f and it's taylor polynomial of degree n, also known as the error between the Taylor polynomial and the function it's trying to approximate. In your case n=2 and a=0, so you just put these values in to the remainder formula you have and work out the 2+1=3rd derivative of your function.

    The p point comes from the mean-value theorem and you don't actually know what it is, just the interval it's on. You said it's on (a,n), did you mean (a,x), for x>a? You can use this to give an upper bound for the error when x>0 by noticing that the third derivative of this function is bounded here.
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