Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 16C (Numerical Analysis)

For fixed $x$ add $K\Pi (t)$ to $p(t)-f(t)$ so the resulting function also vanishes at $t=x$. Repeated Rolle gives a point $\xi$ where its $(n+1)$st derivative vanishes, yielding the formula.

The recurrence $T_{k+1}=2x{T}_k-T_{k-1}$ shows that $T_{n+1}$ has leading coefficient $2^n$. Since its roots are the interpolation nodes, $T_{n+1}=2^n\Pi$, and $|T_{n+1}|\le 1$ on $[-1,1]$.

Since $|f^{(n+1)}|\le k^{n+1}$, the error is at most $k^{n+1}/[2^n(n+1)!]\to 0$ uniformly.

At $x=1$, direct substitution gives a product of odd half-integers. Taking logarithms and applying the hinted lower Riemann-sum estimate yields $\log \Pi (1)>n\log (2/e)$, hence $\Pi (1)>(2/e{)}^n$.