Past exam of the options II course of the University of Oxford / 2007 / P / P3 / c / Solution

Use the corrected interpolant from part (a). The error has $e(x_i)=e^{\prime }(x_i)=0$, and the chosen auxiliary function has $\varphi (x_i)={\varphi }^{\prime }(x_i)=0$ as well as $\varphi (x)=0$. There are $n+2$ distinct zeros of $\varphi$: all the nodes and the additional evaluation point. Rolle's theorem supplies $n+1$ zeros of ${\varphi }^{\prime }$ between these ordered zeros. Together with its $n+1$ nodal zeros, this gives $2n+2$ distinct derivative zeros. Repeated Rolle applications therefore give an interior $\xi$ such that ${\varphi }^{(2n+2)}(\xi )=0$.

The extra claim that $x$ itself must be a zero of ${\varphi }^{\prime }$ is false. For nodes zero and one and $f(t)=t^5$, the Hermite interpolant is $3t^3-2t^2$, so $e(t)=t^2(t-1{)}^2(t+2)$. The auxiliary function is then $\varphi (t)=t^2(t-1{)}^2(t-x)$, and ${\varphi }^{\prime }(x)=x^2(x-1{)}^2\ne 0$ at any interior evaluation point. The corrected zero count above is all that is needed.

Since $h$ has degree at most $2n+1$ and ${\pi }^2$ is monic of degree $2n+2$, differentiating the auxiliary function at the supplied $\xi$ givesHenceAt interpolation nodes the error and the prefactor are zero, so the formula remains valid there as well.