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

The printed product omits the node with index zero. The correct cardinal polynomial isFor example, with two nodes zero and one, values zero and one, and both derivative data zero, the printed index range gives $L_1=1$ and its displayed interpolant is the constant one, which fails the value condition at zero. Thus the source formula cannot be proved literally.

With the corrected product, $L_k(x_j)={\delta }_{kj}$. For $j\ne k$, $L_k^2$ has a double zero at $x_j$, so both basis functions and their derivatives vanish there. At $x_k$, put $\ell =L_k^{\prime }(x_k)$. The value basis $H_k=L_k^2[1-2(x-x_k)\ell ]$ satisfies $H_k(x_k)=1$ and $H_k^{\prime }(x_k)=2\ell -2\ell =0$. The derivative basis $K_k=L_k^2(x-x_k)$ satisfies $K_k(x_k)=0$ and $K_k^{\prime }(x_k)=1$. Both have degree at most $2n+1$. Consequentlyhas every prescribed value and derivative, proving membership in the interpolation class after the required index correction. This corrected Hermite interpolant is used in the following parts.

The difference $d=p-q$ has $d(x_i)=d^{\prime }(x_i)=0$ at every node. Rolle's theorem supplies an additional zero of $d^{\prime }$ in each of the $n$ open intervals between successive nodes. Together with the $n+1$ nodal derivative zeros, these are $2n+1$ distinct zeros of $d^{\prime }$, including the two endpoints. But $d^{\prime }$ is a polynomial of degree at most $2n$, so it is identically zero. Since $d(x_0)=0$, $d$ is also identically zero.

Thus the interpolation polynomial is unique. Once $d^{\prime }=0$ identically, it indeed has arbitrarily many zeros in the open interval as requested. The initial Rolle count naturally includes the endpoints, rather than initially supplying $2n+1$ open-interval zeros. Equivalently, every node is a double root of $d$, so a nonzero difference would be divisible by the degree-$2n+2$ polynomial ${\prod }_i(x-x_i{)}^2$, exceeding its allowed degree.

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.

The cubic Hermite interpolant to the exponential at zero and one isIts endpoint values are one and $e$, and its endpoint derivatives are also one and $e$. The remainder formula gives $e^x-h(x)=e^{\xi }{x}^2(x-1{)}^2/24$ for an interior $\xi$. On this interval $e^{\xi }\le e$ and $x(1-x)\le 1/4$, so