Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 14B / d / Solution

Continue with $D,D_i$ from part (c). Comparing omission of $x_i$ and $x_n$ givesComparing omission of $x_0$ and $x_n$ gives $D_0=D_n+(x_n-x_0)D$. Eliminating $D$ provesHere $D_n=f[x_0,\dots ,x_{n-1}]$ and $D_0=f[x_1,\dots ,x_n]$, so this is exactly the required omitted-node identities for divided differences. The original nodes need not be ordered, and the formula remains valid even if $\gamma$ lies outside $[0,1]$.