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

For the natural logarithm, $f(1)=0$, $f^{\prime }(1)=1$, $f^{\prime \prime }(1)=-1$, and $f^{(3)}(x)=2/x^3$. The quadratic Taylor polynomial therefore givesThe Taylor remainder is $R_3=2(0.1{)}^3/(6{\xi }^3)$ with $1<\xi <1.1$. It is positive, so the approximation is an underestimate. Moreover,Thus $1/3000\approx 0.0003334$ is a valid upper bound on the absolute error.