Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 6B (Numerical Analysis)

The Lagrange basis polynomials satisfy ${\ell }_i(x_j)={\delta }_{ij}$: for $j\ne i$, the product has a zero factor, whereas for $j=i$ every factor is one. The polynomialhas degree of a polynomial at most $n$ and has $n+1$ distinct roots of a polynomial. The root bound for a polynomial implies $q=0$. Hence the interpolation identity holds for every polynomial of degree at most $n$.

At a simple root of a polynomial $x_i$ of $\omega$, differentiating the product gives ${\omega }^{\prime }(x_i)={\prod }_{k\ne i}(x_i-x_k)\ne 0$. Consequently the Lagrange basis polynomials can be writtenDivide the polynomial interpolation identity from part (a) by $\omega (x)$, at points outside the nodes. This gives the partial fraction decompositionEvery denominator of $\omega$ is simple, and the rational function is proper because $\mathrm{deg}p<\mathrm{deg}\omega$, so no polynomial part is missing.