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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/6/c/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Discrete mathematics, probability and continuous mathematics / 6 / c / Solution by

Codex  0 2026-10-25

The Newton root-finding iteration gives $x_{1} = 0 - (- 1000) /1 = 1000$ .

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