Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / P (Numerical Analysis)

Use Gaussian elimination with partial pivoting, choosing the largest absolute available entry in the current pivot column. The first pivot is 2, with multipliers $1/2,1/2,1/5$. After eliminating that column, the augmented system isThe second pivot is $-1$, already largest in its remaining column. Its multipliers are zero and $2/5$, giving row four $(0,0,3,-2\mid 0)$. In column three, partial pivoting now requires exchanging rows three and four, since $3>3/4$. The final elimination multiplier is $-1/4$. ThusBack substitution gives $\boxed{(w,x,y,z)=(1,-1,0,0)}$.

The permutation and recorded multipliers, with the earlier multiplier rows swapped at the third pivot, areDirect multiplication verifies $PA=LU$. Since $U$ is invertible and $A=P^{T}LU$,The two matrices are similar, and similar matrices have the same characteristic polynomial and eigenvalues.

Let $M_j=s^{\prime \prime }(j)$. A natural cubic spline has $M_0=M_2=0$. The interior spline equation on equal unit intervals isso $M_1=6$. The standard second-derivative interpolation formula, or direct integration of the linear second derivative, givesThe two pieces have equal value, first derivative, and second derivative at one, and $s^{\prime \prime }$ vanishes at zero and two.

Adding the linear function $x$ changes the nodal data by $(0,1,2)$ without affecting second derivatives. By uniqueness, the second requested natural cubic spline is $s(x)+x$, with pieces $x^3$ and $-x^3+6x^2-6x+2$ respectively.