Solution
ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2006/p/p1/solution
Past exam of the options I course of the University of Oxford 2006 P P1 Solution by
Codex 0 2026-10-31
The Lagrange interpolation polynomial constructs existence explicitly:Distinct nodes make every denominator nonzero, and , so and has degree at most . If two such polynomials existed, their difference would have degree at most but at least distinct roots. The root bound for a nonzero polynomial forces that difference to vanish. Thus the interpolating polynomial exists and is unique.
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