Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 15

With the first $n$ terms counted from the constant term, Taylor's theorem giveswhere $\xi$ lies between $a$ and $x$. Continuous derivatives through order $n$ on the joining interval are a sufficient hypothesis for this Taylor theorem with Lagrange remainder. The Taylor remainder measures the error of the finite Taylor polynomial; convergence of the corresponding infinite Taylor series is a separate question.

For $f(x)=x^{1/3}$, the required derivatives areAt $a=8$ these give $f(8)=2$, $f^{\prime }(8)=1/12$, and $f^{\prime \prime }(8)=-1/144$. Setting $x-a=1$, the three-term Taylor polynomial yieldsBy the Taylor theorem with Lagrange remainder, the signed error isIt is positive, and ${\xi }^{-8/3}<8^{-8/3}=1/256$. ConsequentlyThe last comparison is exact because $5\cdot 4000=20000<20736$. Thus the estimate is an underestimate with the required rigorous absolute error bound.

Use the Maclaurin series $\sin x=x-x^3/6+O(x^5)$ and $e^u=1+u+u^2/2+u^3/6+u^4/24+O(u^5)$. The terms through fourth degree giveIn particular the cubic coefficient cancels; it has not been omitted accidentally.

The Maclaurin series give $e^{-x^3}=1-x^3+O(x^6)$ and $\cosh x=1+x^2/2+x^4/24+O(x^6)$. Inverting the latter using the geometric series givesMultiplying the two expansions yieldsThe remainder is $O(x^5)$ because the product already generates an $x^5$ term from $(-x^3)(-x^2/2)$.