Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 5 / a / iii / Solution

A small difference between successive secant method iterates alone does not certify a small absolute error. Instead use a residual root error bound. On $[2,3]$,and $f(2)<0<f(3)$, so there is a unique root $r\in (2,3)$. For an iterate $x\in [2,3]$, the mean value theorem givesTherefore stop when $x\in [2,3]$ and $|x{\log }_{2}x-4|\le (1+1/\ln 2)10^{-10}$. This guarantees the desired absolute error in exact arithmetic; the evaluated residual of a root equation should include an allowance for its much smaller roundoff error. A safeguarded implementation can retain the bracket $[2,3]$ and use bisection if a proposed secant method step leaves it.