 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability/continuous-mathematics/6/d/i/solution

Δ = H (f)^{- 1} \nabla f

. Its second row equation is

- 4 Δ_{x} + 4 Δ_{y} = 4 y - 4 x

, so

Δ_{y} = Δ_{x} + y - x

. Therefore

y_{n + 1} = y_{n} - Δ_{y} = x_{n} - Δ_{x} = x_{n + 1}

. The given first coordinate yields

φ (x, y) = \frac{2 x ^{3}}{3 x ^{2} - 1} .

The Newton step is defined where

det H (f) = 16 (3 x^{2} - 1) \neq = 0

; both coordinates after any defined step are equal.

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