Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 17

A cubic version of Taylor's theorem requires the corresponding higher differentiability, rather than only one derivative. For example, if $f$ has four continuous derivatives on the interval between $a$ and $x$, thenwhere the Lagrange form of the Taylor remainder isfor some $\xi$ strictly between $a$ and $x$ when they differ. At $x=a$, the remainder is zero. These hypotheses apply to both functions used below.

The required derivatives are $f^{\prime }(x)=1/(2\sqrt{x})$, $f^{\prime \prime }(x)=-1/(4x^{3/2})$, $f^{(3)}(x)=3/(8x^{5/2})$, and $f^{(4)}(x)=-15/(16x^{7/2})$. The four-term Taylor polynomial about one, evaluated at two, isThe Taylor remainder is $R_3(2)=-5/(128{\xi }^{7/2})$ for some $1<\xi <2$. ThusIts negative sign also shows that this approximation is an overestimate. Bounding the fourth derivative throughout the interval establishes the requested error estimate without using a decimal value of $\sqrt{2}$.

Factor $1+e^x=2e^{x/2}\cosh (x/2)$. This givesThe last function is even. Its Taylor series begins $\ln \cosh (x/2)=x^2/8+O(x^4)$, since $\cosh (x/2)=1+x^2/8+O(x^4)$ and $\ln (1+w)=w+O(w^2)$. ThereforeIn particular, the cubic coefficient is zero; the identity explains its cancellation before any lengthy differentiation.