 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ib/paper-2/14b/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

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

D_{i} = γ D_{n} + (1 - γ) D_{0}, γ = \frac{x _{i} - x _{0}}{x _{n} - x _{0}} .

Here

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

γ

lies outside

[0, 1]

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