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

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.

For $d=p-q$, all $d(x_i)$ and $d^{\prime }(x_i)$ vanish. Between each consecutive pair of nodes, Rolle's theorem gives one additional zero of $d^{\prime }$. Along with the $n+1$ nodal zeros, this supplies $2n+1$ distinct zeros on the closed interval. Since $d^{\prime }$ is a polynomial of degree at most $2n$, it must be the zero polynomial. Thus $d$ is constant, and $d(x_0)=0$ makes it zero. Hence the interpolating polynomial is unique.

The direct zero count includes the two endpoints; its guaranteed interior count before the degree argument is $2n-1$. After $d^{\prime }=0$ is proved, $d^{\prime }$ vanishes at every point of the open interval, so the stronger interior-zero assertion in the source follows as well. Another immediate uniqueness proof is that $d$ has a double zero at every node, so is divisible by ${\prod }_i(x-x_i{)}^2$ of degree $2n+2$, although its own degree is at most $2n+1$.

Use the corrected Hermite interpolating polynomial from part (a). The auxiliary function $\phi$ vanishes at all $n+1$ nodes and at the additional point $x$. Its derivative vanishes at all nodes, because both $e^{\prime }(x_i)$ and the derivative of ${\pi }^2$ vanish there. Rolle's theorem between the $n+2$ ordered zeros of $\phi$ gives $n+1$ further, distinct zeros of ${\phi }^{\prime }$. Thus ${\phi }^{\prime }$ has $2n+2$ distinct zeros on the closed interval. The printed phrase that these include $x$ is not generally true; the extra zeros lie in intervals between zeros of $\phi$, and ${\phi }^{\prime }(x)$ need not vanish.

Repeated application of Rolle's theorem gives ${\phi }^{(2n+2)}(\xi )=0$ for some interior $\xi$. Since $h$ has degree at most $2n+1$ and ${\pi }^2$ is monic of degree $2n+2$,Rearranging givesAt an interpolation node, both sides are zero, so the error formula extends to the whole closed interval. The false location claim about one zero is unnecessary for this result.

For nodes $0,1$ and $f(x)=e^x$, the cubic Hermite interpolating polynomial isIndeed its values at zero and one are $1,e$, and its derivatives there are also $1,e$. The Hermite interpolation error formula givessince $0\le \xi \le 1$ and $x(1-x)\le 1/4$ on $[0,1]$.