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Math Help - Taylor's theorem

  1. #1
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    Taylor's theorem

    Suppose that f(x)=\ln (1+x)

    1) Express f(x) in the form p_1(x)+R_2(x), where p_1 is the first Taylor polynomial for f about 0 and R_2 is the Lagrange formula for the remainder.

    f(x)=p_1(x)+R_2(x)

    p_1(x)=f(0)+f'(0)x

    \Longrightarrow p_1(x)=x

    R_2(x)=\frac{f''(c)}{2!}x^2

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

    f(x)=x-\frac{x^2}{2(1+c)^2}

    2) Suppose that x\in [-0.1, 0.1] and consider the approximation \ln (1+x) \approx x. Use the answer in the first part to show that an upper bound for the absolute error in this approximation is \frac{1}{162}.

    |Error|=|f(x)-p_1(x)|

    =|R_2(x)|

    How would I do this question?
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  2. #2
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    Quote Originally Posted by acevipa View Post
    Suppose that f(x)=\ln (1+x)

    1) Express f(x) in the form p_1(x)+R_2(x), where p_1 is the first Taylor polynomial for f about 0 and R_2 is the Lagrange formula for the remainder.

    f(x)=p_1(x)+R_2(x)

    p_1(x)=f(0)+f'(0)x

    \Longrightarrow p_1(x)=x

    R_2(x)=\frac{f''(c)}{2!}x^2

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

    f(x)=x-\frac{x^2}{2(1+c)^2}

    2) Suppose that x\in [-0.1, 0.1] and consider the approximation \ln (1+x) \approx x. Use the answer in the first part to show that an upper bound for the absolute error in this approximation is \frac{1}{162}.

    |Error|=|f(x)-p_1(x)|

    =|R_2(x)|

    How would I do this question?
    Well, you have already determined that R_2(x)= -\frac{x^2}{(1+ c)^2} for some c between -0.1 and 0.1 so that |R_2(x)|= \frac{x^2}{(1+ c)^2}. Now what is the largest possible value for that? Remember that a fraction achieves it maximum value when the numerator is maximum and the denominator is minimum.
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