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

A Givens rotation equals the identity except in coordinates $i,j$, where its block can be taken asThe two block columns are orthonormal because $c^2+s^2=1$; other coordinates are unchanged, so the full matrix is orthogonal. If $(v_i,v_j)\ne (0,0)$, take $r=\sqrt{v_i^2+v_j^2}$, $c=v_i/r$ and $s=v_j/r$. The transformed pair is $(r,0)$. If both entries are zero, the identity already suffices.

For the specified vector one possible matrix isIts rows are orthonormal. It is not unique: premultiply by any orthogonal transformation fixing the second coordinate axis pointwise. This rotates or reflects the other two output coordinates without changing the required image. Equivalently its second row must be $v^T/\Vert v\Vert$, but any orthonormal completion of that row works.

Richardson extrapolation cancels a leading discretization error by combining two mesh sizes. If $A(h)=I+c{h}^p+O(h^q)$ with $q>p$, thenUse a uniform grid $x_i=a+ih$, $0\le i\le n$, with even $n$. The composite trapezium rule isThe composite Simpson rule, formed by integrating a quadratic interpolant on each pair of successive panels, isAssume an even-power trapezium error expansion $T_h=I+c_2{h}^2+O(h^4)$; continuous fourth derivative is a standard sufficient smoothness hypothesis. The coarse grid has step $2h$, so one extrapolation step is $(4T_h-T_{2h})/3$. Its endpoint coefficients are $h/3$, its odd-index coefficients are $4h/3$, and its interior even-index coefficients are $2h/3$. ThereforeThe algebraic identity between the two rules holds for any sampled values; the smoothness and error assumptions are used to infer the improved accuracy.