 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-resit/5/a/iii/solution

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.

 New to topics? Read the docs here!