Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics and probability resit / 6

The displayed rational expression has domain $\mathbb{R}\setminus \{0\}$; its value at zero is undefined despite the printed $\mathbb{R}\to \mathbb{R}$ notation. On that domain, the power rule givesThese derivatives are smooth on every interval avoiding zero, including all intervals used below.

Substitution givesIn particular, $1$ is a stationary point. For $x$ sufficiently close to $1$, Taylor's theorem with fifth-order remainder giveswhere $\xi$ lies between $1$ and $x$. The fifth derivative is bounded near $1$, so the remainder is $O((x-1{)}^5)$ and is smaller in magnitude than half the positive quartic term for sufficiently small nonzero $x-1$. Therefore $x=1$ is a strict local minimum, despite the vanishing second derivative.

An exact algebraic check gives the stronger global conclusion:Equality holds only at $x=1$, so it is the unique global minimum on the actual domain.

Let $h=(b-a)/n$ and let $M_4={\max }_{x\in [a,b]}|f^{(4)}(x)|$. A composite Simpson's rule panel covers two strips, so it has width $2h$ and absolute error at mostThere are $n/2$ panels. By the triangle inequality, the absolute error of their sum is at most the sum of their absolute errors. HenceThis assumes even $n$ and, for the usual Simpson remainder theorem, $f\in C^4[a,b]$. In particular its fourth derivative is bounded on the interval. It is the error bound for composite Simpson's rule, expressed in strip count rather than panel count.^[ref]

Write $f^{(4)}(x)=24(5-4x)/x^6$. Its derivative is $f^{(5)}(x)=240(2x-3)/x^7$, so on $[1,2]$ it decreases to $x=3/2$ and then increases. Its endpoint and turning-point values areThus $M_4=24$, and the preceding bound becomesFor the rounding estimate, standard binary64 has spacing $ϵ=2^{-52}$ above one. Balancing this nominal absolute accuracy with the bound givesThus a few thousand even strips, roughly $5000$, reach the double-precision scale according to this bound. A scale-aware estimate leads to the same order. Indeed,lies in magnitude between $4$ and $8$, where the binary64 spacing is $2^{-50}$. Balancing the error bound against half that spacing gives about $4.16\times 10^3$ strips.

These are saturation estimates from balancing quadrature truncation and round-off error, not sharp guarantees of the actual rounding plateau. The error bound can be conservative, and evaluation and summation errors can limit accuracy earlier. Merely storing the final result does not specify every source of rounding error.

Not near the minimum at $1$. The derivative being set to zero factors asso $1$ has multiplicity of a root equal to three. The ordinary Newton iteration at a multiple root is generally only linearly convergent, with error ratio $(3-1)/3=2/3$ here.

In fact, cancellation in $x-f^{\prime }(x)/f^{\prime \prime }(x)$ gives the exact error identity, away from the original undefined quotient at the root,For iterates approaching $1$, the factor tends to $2/3$, proving linear convergence. The other zero of $f^{\prime }$ at $-1$ is simple and could instead give quadratic local convergence; the failure at $1$ is due to its multiplicity, not to every stationary point of this rational function.