Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics IV / 7 / a

For $0\le x\le 7$, we have $2\le 2+x\le 9$. Taking the nonnegative square root givesThe derivative issoIn particular, the mean value theorem makes $G$ a contraction mapping on this interval.

Set $H(x)=g(x)-x$. This function is continuous on $[a,b]$. Since $g$ takes values in $[a,b]$,The intermediate value theorem therefore supplies $\xi \in [a,b]$ with $H(\xi )=0$, equivalently $g(\xi )=\xi$. This includes the possibility that one endpoint is already a fixed point.

For uniqueness, the mean value theorem gives, for any $x,y\in [a,b]$,If both $\xi$ and $\eta$ are fixed points, thenSince $\gamma <1$, this forces $|\xi -\eta |=0$. Thus there exists exactly one fixed point. When $a=b$, the conclusion follows immediately from $g(a)=a$; otherwise the argument above applies. The contraction mapping estimate also proves convergence of the associated fixed-point iteration, as used next.

The fixed-point iteration stays inside $[0,7]$ because $G$ maps that interval to itself. A fixed point satisfies $\xi =\sqrt{2+\xi }$, so $\xi \ge 0$ andOnly $\xi =2$ is admissible; $-1$ fails the unsquared equation. With $q=1/(2\sqrt{2})$, the mean value theorem gives the error boundInduction yields $|x_k-2|\le q^k|x_0-2|\le 5q^k\to 0$, since $0<q<1$. Therefore, for every allowed starting value,This estimate proves convergence directly, without assuming in advance that the sequence has a limit.