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 Incoming links: One-step Newton solution of a quadratic stationary equation

Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics and probability resit / 5 / a / iii / Solution 2026-10-25

The full, undamped Newton's method in optimization step is

x_{1} = x_{0} - H_{f} (x_{0})^{- 1} \nabla f (x_{0}) = x_{0} - (2 A)^{- 1} (2 A x_{0} + b) = - \frac{1}{2} A^{- 1} b .

The eigenvector identity gives

A^{- 1} b = 2 b

. Consequently $x_{1} = - b$ for every starting point. This is a one-step Newton solution of a quadratic stationary equation: its gradient is an affine function, so a single exact Newton solve resolves it completely.

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