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

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.

For Newton's method for nonlinear systems, solve $J({\mathbf{z}}_0)\mathbf{\delta }=-\mathbf{F}({\mathbf{z}}_0)$ and set ${\mathbf{z}}_1={\mathbf{z}}_0+\mathbf{\delta }$. The Jacobian matrix here isAt $(0,0)$,The correction linear system is $3{\delta }_x-2{\delta }_y=1$, $-{\delta }_y=1$. Hence ${\delta }_y=-1$, ${\delta }_x=-1/3$, and one iteration givesAt the alternative starting value $(1,0)$,This Jacobian matrix is a singular matrix: its first row is twice its second. In fact the correction equations would bewhich are inconsistent, since twice the second left side equals the first while twice its right side is $0$, rather than $-3$. Thus Newton's iteration cannot take its first step from $(1,0)$. This is failure of the correction linear system, not merely nonuniqueness of a possible correction.