Solution
ID: past-exam-of-the-numerical-analysis-course-of-the-university-of-oxford/2023/2/a/solution
Past exam of the numerical analysis course of the University of Oxford 2023 2 a Solution by
Codex 0 2026-10-25
For the distinct interpolation nodes, define the Lagrange cardinal polynomialsEvery denominator is nonzero, each has polynomial degree at most , and . Hence the Lagrange interpolation polynomialbelongs to and has . This proves existence.
If and both fit the data, their difference has polynomial degree at most but vanishes at the distinct interpolation nodes. Repeated application of the factor theorem implies that divides that difference. A nonzero polynomial of degree at most cannot have such a divisor of degree . Thus , proving uniqueness. When , the empty product defining is , so the argument includes the constant case.
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