 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/15/b/solution

f (1) = 0

f^{'} (1) = 1

f^{''} (1) = - 1

, and

f^{(3)} (x) = 2/ x^{3}

. The quadratic Taylor polynomial therefore gives

ln (1.1) \approx 0.1 - \frac{0. 1 ^{2}}{2} = 0.095 .

R_{3} = 2 (0.1)^{3} / (6 ξ^{3})

with

1 < ξ < 1.1

. It is positive, so the approximation is an underestimate. Moreover,

0 < ln (1.1) - 0.095 < \frac{( 0.1 ) ^{3}}{3} = \frac{1}{3000} .

Thus

1/3000 \approx 0.0003334

is a valid upper bound on the absolute error.

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