Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 3G (Analysis II)

A contraction mapping on a metric space $(X,d)$ is a map $f:X\to X$ for which there is a single constant $q$ with $0\le q<1$ such thatThe same $q$ must work for every pair. Merely decreasing each nonzero distance strictly does not suffice for this definition.

The contraction mapping theorem states that a contraction mapping of a nonempty complete metric space into itself has a unique fixed point $x_{\ast }$. For any starting point $x_0$, the iterates $x_{n+1}=f(x_n)$ converge to $x_{\ast }$, with the quantitative estimateCompleteness and the self-map condition are essential: the geometric estimate makes the iterates a Cauchy sequence, completeness supplies their limit in $X$, and continuity of the contraction mapping makes that limit a fixed point. If two fixed points existed, their distance would be at most $q$ times itself, forcing equality of the points.

Define $F_i(x)=a_i+\frac{1}{6}{\sum }_{j\ne i}\sin x_j$. Solving the system is equivalent to finding a fixed point of $F:{\mathbb{R}}^3\to {\mathbb{R}}^3$. The mean value theorem implies $|\sin u-\sin v|\le |u-v|$. In the specified metric,Each coordinate difference occurs exactly twice. This metric makes ${\mathbb{R}}^3$ a nonempty complete metric space, because a Cauchy sequence in it is Cauchy in each real coordinate, and coordinate convergence implies convergence in the sum metric. Thus $F$ is a contraction mapping with constant $1/3$.

There is exactly one solution for every choice of $(a_1,a_2,a_3)$, by the contraction mapping theorem. Iteration of $F$ from any starting vector also constructs it.