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

The secant method starts from two approximations $x_0,x_1$. Replace the function near its zero of a function by the secant line through $(x_{k-1},f(x_{k-1}))$ and $(x_k,f(x_k))$; the zero of that line is the next approximation:Stop immediately if $f(x_k)=0$; otherwise the denominator must be nonzero. Each further step needs one new function evaluation and no derivative. There is no automatic bracketing or global convergence guarantee.^[ref]

Use $f(x)=x{\log }_{2}x-4$. Then $f(1)=-4$, $f(2)=-2$, and the secant method givesWriting $L={\log }_{2}3$, the next step isThus the requested approximation is $x_3=2.73$ to three significant figures. All intermediate logarithms should be retained at higher precision before this final rounding.

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.