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Taylor Polynomial

  1. Sep 24, 2005 #1
    Find the thrid taylor polynomial P3(x) for the function [itex] f(x) = \sqrt{x+1} [/itex] about a=0. Approximate f(0.5) using P3(x) and find actual error

    thus Maclaurin series

    [tex] f(x) = f(0) + f'(0)x + \frac{f''(0)}{2} x^2 + \frac{f^{3}(0)}{6} x^3 [/tex]

    [tex] f(x) = x + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{3}{48} x^3 [/tex]
    am i right so far?
    To approximate f(0.5) i simply put x=0.5 in the above equation?
    How do i fin the actual error, though?
    DO i have to use the remainder in this? Please help!

    Thank you
  2. jcsd
  3. Sep 24, 2005 #2

    Tom Mattson

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    All but the first term is right. [itex]f(0)\neq x[/itex]

    After you fix it, yes.

    Plug x=0.5 into f(x) on a calculator, and subtract your result from it. You won't exactly get the "actual" error because your calculator approximates, too. But it will be a very good estimate.

    That depends on what is asked for. The remainder doesn't give you the actual error, but rather the maximum of the actual error. So unless you were asked to put bounds on the error, I would think that you would not have to use the remainder.
  4. Sep 24, 2005 #3
    [tex] f(x) = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{3}{48} x^3 [/tex]

    i see the problem, its fixed now :smile:

    im being cautious so im goingto put hte upper limits

    [tex] R_{4} = \frac{15}{384} (c+1)^{\frac{-7}{2}} x^4[/tex]

    so the error must be lesser than or equal to this R4 value. THat c value lies between 0.5 and x?

    Is this right?
  5. Sep 26, 2005 #4
    is this how one would solve for the maximum possible error as stated in the above post? Please do advise

    Thank you for your help and input
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