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Math Help - Error in taylor polynomial

  1. #1
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    Error in taylor polynomial

    Suppose that f(x)=\sqrt{1+x} and let p_2 denote the second Taylor polynomial for f about 0. If x\in [0,1] then show that the absolute error in the approximation f(x)\approx p_2(x) does not exceed \frac{1}{16}

    This is what I've done so far:

    f(x)=\sqrt{1+x}\Rightarrow f(0)=1

    f'(x)=\frac{1}{2}(1+x)^{-\frac{1}{2}}\Rightarrow f'(0)=\frac{1}{2}

    f''(x)=-\frac{1}{4}(1+x)^{-\frac{3}{2}}\Rightarrow f''(0)=-\frac{1}{4}

    p_2(x)=1+\frac{1}{2}x-\frac{1}{8}x^2

    What should I do next? I'm thinking of using the Lagrange formula, but not too sure how I would use it
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by acevipa View Post
    Suppose that f(x)=\sqrt{1+x} and let p_2 denote the second Taylor polynomial for f about 0. If x\in [0,1] then show that the absolute error in the approximation f(x)\approx p_2(x) does not exceed \frac{1}{16}

    This is what I've done so far:

    f(x)=\sqrt{1+x}\Rightarrow f(0)=1

    f'(x)=\frac{1}{2}(1+x)^{-\frac{1}{2}}\Rightarrow f'(0)=\frac{1}{2}

    f''(x)=-\frac{1}{4}(1+x)^{-\frac{3}{2}}\Rightarrow f''(0)=-\frac{1}{4}

    p_2(x)=1+\frac{1}{2}x-\frac{1}{8}x^2

    What should I do next? I'm thinking of using the Lagrange formula, but not too sure how I would use it
    The Lagrange form of the remainder is:

    R_N=\dfrac{f^{(3)}(\xi)x^3}{3!}

    for some \xi in [0,1].

    So now find an upper bound for |R_n|

    CB
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