Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics IV / 7 / b

To find a critical point, apply multivariable Newton iteration to the gradient, rather than to the scalar value $F$. The gradient and Hessian matrix areWrite the correction as ${\mathbf{s}}_0={\mathbf{x}}_1-{\mathbf{x}}_0$. At ${\mathbf{x}}_0=(-1,-1)$, Newton's method in optimization requiresSubtracting the second equation from the first gives $10s_y=7$, so $s_y=7/10$ and $s_x=13/10$. HenceThe gradient there isA critical point has not been found after one iteration.

At the alternative starting point $(0,2/3)$, the Hessian matrix and gradient areThe correction would need to satisfy $2s_x+2s_y=-4/3$ and $2s_x+2s_y=0$ simultaneously. This linear system is inconsistent. The standard iteration cannot take even its first step: the Hessian matrix is a singular matrix, and the required correction does not exist. This is a Newton correction compatibility condition failure, not merely an ambiguity between several corrections.