Induct on $k$, using the recursive divided-difference formula. After substituting the two induction hypotheses, use
$$
f[t_1,\ldots,t_m]-f[t_0,\ldots,t_{m-1}]
=(t_m-t_0)f[t_0,\ldots,t_m]
$$
and the analogous identity for $g$; adjacent terms telescope. This proves
$$
\boxed{(fg)[t_0,\ldots,t_k]
=\sum_{m=0}^kf[t_0,\ldots,t_m]g[t_m,\ldots,t_k]}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 318 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-318/3/c/solution