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Math Help - Taylor Series cubic approximation

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    Taylor Series cubic approximation

    Using Taylor's formula with a=0 and n=3, how would you find the cubic approximation of f(x)= 1/(1-x) at x=0, given the upper bound for the magnitude of the error in the approximation when lxl< 0.1 ?
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    Quote Originally Posted by TAG16 View Post
    Using Taylor's formula with a=0 and n=3, how would you find the cubic approximation of f(x)= 1/(1-x) at x=0, given the upper bound for the magnitude of the error in the approximation when lxl< 0.1 ?
    The derivatives are

     f'(x) = \frac{1}{(1-x)^2},\;\; f'(0) = 1
     f''(x) = \frac{2}{(1-x)^3},\;\; f''(0) = 2!
     f'''(x) = \frac{3!}{(1-x)^4},\;\; f'''(0) = 3!
     f^{(4)}(x) = \frac{4!}{(1-x)^5},\;\; \text{for the remainder}
    The Taylor polynomial P_3(x) = 1 + x + x^2 + x^3 and the Taylor remainder R_3(x) = f^{(4)}(c) \frac{x^4}{4!} = \frac{x^4}{(1-c)^5} where c is between 0 and x. Since |x| < 0.1 then we are trying to determine the maximum of |R_3(x)| = \left|\frac{x^4}{(1-c)^5}\right| on the domain [-0.1,0.1]\;\times\;[-0.1,0.1] and so this would occur at x = \pm 0.1,\;\;c = 0.1 and the maximum error would be

    |R_3(x)| \le 0.00006209.
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