Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / P / 3 / a / Solution

The printed product omits the node $x_0$: its lower index must be zero for the stated formula to hold. For example, with $n=1$, nodes $0,1$, and data $f_0=0,f_1=1,g_0=g_1=0$, the printed $L_{1,1}=1$ gives $h=1$, which fails the first value condition. Use the corrected Lagrange basis polynomialsThey satisfy $L_k(x_j)={\delta }_{kj}$. Define $H_k=L_k^2[1-2(x-x_k)L_k^{\prime }(x_k)]$ and $K_k=L_k^2(x-x_k)$. At a node with $j\ne k$, the double zero of $L_k^2$ gives $H_k(x_j)=H_k^{\prime }(x_j)=K_k(x_j)=K_k^{\prime }(x_j)=0$. At $x_k$, direct differentiation givesBoth polynomials have degree at most $2n+1$. Consequently $h={\sum }_k(f_k{H}_k+g_k{K}_k)$ has the specified values and derivatives, and is the required Hermite interpolating polynomial. The source product is retained literally in the private transcription; this solution identifies its necessary correction.