Past exam of the probability, statistics and options I course of the University of Oxford / 2007 / P (Numerical Analysis)

A Givens rotation acts as the identity except on coordinates $i,j$, where its block can be writtenThe block has orthonormal columns because ${\cos }^2\theta +{\sin }^2\theta =1$ and their inner product is zero, so $J^{T}J=I$. For $r=\sqrt{v_i^2+v_j^2}>0$, choose $\cos \theta =v_i/r$, $\sin \theta =v_j/r$. The new coordinates are $(r,0)$, eliminating the $j$th entry. If both entries are zero, the identity already works.

For the specified vector, one suitable matrix isIts rows are orthonormal, so it is orthogonal; in fact its determinant is one. The positive nonzero output component must be five because an orthogonal matrix preserves Euclidean norm.

The matrix is not unique. Left multiplication by any orthogonal matrix fixing the second coordinate vector preserves this output. For example arbitrary rotations of the first and third coordinates provide infinitely many alternatives.

Richardson extrapolation combines approximations with a known leading error power to cancel that error. If $A(h)=I+c{h}^p+O(h^{p+q})$, then $(2^{p}A(h)-A(2h))/(2^p-1)$ removes $c{h}^p$.

Let $x_j=a+jh$ for an even number $n$ of equal subintervals, with $x_n=b$. The composite trapezoidal rule isThe Simpson's rule isAssume an error expansion $T_h=I+c_2{h}^2+O(h^4)$, with the same coefficient $c_2$ on refined uniform meshes; $f\in C^4([a,b])$ is sufficient for this form. The coarser rule is $T_{2h}=h[f_0+f_n+2{\sum }_{\mathrm{even}}{f}_j]$. Direct coefficient collection givesThe leading $h^2$ errors cancel, leaving $S_h=I+O(h^4)$. The identity of the numerical integration weights is exact; the error-order conclusion additionally uses the smoothness and even-power error assumptions.