Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 2F / Solution

A map $h:X\to X$ is a contraction if there is a constant $q<1$ such thatfor every $x,y\in X$.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point, and that the iterates from every starting point converge to it. To prove this, choose $x_0\in X$ and put $x_{n+1}=h(x_n)$. Thenso, for $m>n$,Thus $(x_n)$ is Cauchy and converges, by completeness, to some $x_{\ast }$. A contraction is continuous, soIf $y_{\ast }$ is another fixed point, thenforcing $x_{\ast }=y_{\ast }$.

For the Newton mapone has $g(r)=r$ andOn the given neighbourhood,Since $f(r)=0$, choose a closed interval $U^{\prime }$ centred at $r$ and contained in $U$ so small that $M|f(x)|/{\delta }^2\le 1/2$ there. Thenso $g(U^{\prime })\subseteq U^{\prime }$ and $g$ is a contraction on the complete interval $U^{\prime }$. The local contraction proof for Newton iteration therefore shows that $r$ is the unique fixed point of $g$ on $U^{\prime }$.

Solved by gpt-5.6-sol high.