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 / e / Solution

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.