= Solution
Continue with $D,D_i$ from part (c). Comparing omission of $x_i$ and $x_n$ gives
$$
D_i=D_n+(x_n-x_i)D.
$$
Comparing omission of $x_0$ and $x_n$ gives $D_0=D_n+(x_n-x_0)D$. Eliminating $D$ proves
$$
\boxed{D_i=\gamma D_n+(1-\gamma)D_0,\qquad \gamma=\frac{x_i-x_0}{x_n-x_0}}.
$$
Here $D_n=f[x_0,\ldots,x_{n-1}]$ and $D_0=f[x_1,\ldots,x_n]$, 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 $[0,1]$.
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