Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / P / P3

Interpolation means $p_n(x_i)=f(x_i)$ for every node. Define the cardinal polynomials ${\ell }_i(x)={\prod }_{j\ne i}(x-x_j)/(x_i-x_j)$, which are well defined because the nodes are distinct. They have degree $n$ and satisfy ${\ell }_i(x_j)={\delta }_{ij}$. Hencebelongs to ${\Pi }_n$ and interpolates. If two polynomials in ${\Pi }_n$ interpolate the same data, their difference has $n+1$ distinct zeros but degree at most $n$, so it is zero. This proves uniqueness. The phrase “of degree $n$” denotes membership in ${\Pi }_n$ here: the actual degree may be smaller, for example for constant data.

Multiplying the Vandermonde matrix $A_n$ by a coefficient vector $(c_0,\dots ,c_n{)}^T$ evaluates the polynomial ${\sum }_j{c}_j{x}^j$ at all nodes. If the result is zero, part (a)'s uniqueness for the zero interpolation data forces every coefficient to be zero. Thus the square matrix has trivial kernel and is invertible. Equivalently the interpolation map is a linear bijection between two spaces of dimension $n+1$.

If $x$ is a node, both sides vanish and any $\xi \in (a,b)$ works. Otherwise put $\omega (t)={\prod }_i(t-x_i)$ and $C=(f(x)-p_n(x))/\omega (x)$. The function $H(t)=f(t)-p_n(t)-C\omega (t)$ has $n+2$ distinct zeros, at the nodes and at $x$. Repeated application of Rolle's theorem gives an interior point $\xi$ where $H^{(n+1)}(\xi )=0$. As $p_n^{(n+1)}=0$ and ${\omega }^{(n+1)}=(n+1)!$, this gives $C=f^{(n+1)}(\xi )/(n+1)!$. ThereforeThe distinct nodes and $n\ge 1$ ensure $a<b$, so the interior point is meaningful.

Every derivative of order $n+1$ of $\sin 2x$ has magnitude at most $2^{n+1}$. Pair the symmetric nodes $a=i/n$ and $1-a$. For $0\le x,a\le 1$,If there is an unpaired middle node, its factor has magnitude at most $1/2$. Thus the full nodal product is at most $2^{-(n+1)}$, giving the stronger bound $\Vert f-p_n{\Vert }_{\infty }\le 1/(n+1)!$. In particular this implies the requestedThe bound tends to zero by factorial growth, so $\boxed{p_n\to f\text{ uniformly on }[0,1]}$. This conclusion uses smoothness of this particular function and does not assert uniform convergence of equispaced interpolation for arbitrary continuous data.