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

The greatest common divisor of $54$ and $21$ is $3$, which divides $906$. Divide the linear Diophantine equation by $3$ to obtainReducing modulo $7$ gives $4x\equiv 1(\mathrm{mod}7)$, so $x\equiv 2(\mathrm{mod}7)$. Substitution gives the complete family of integer solutionsEvery such pair satisfies the equation, and every solution must have this residue for $x$, so none are omitted. Positivity requires $2+7t>0$ and $38-18t>0$. For integer $t$ these are exactly $0\le t\le 2$. Thus the pairs with both entries positive are

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.

For multivariable Newton iteration, use the Jacobian matrixAt ${\mathbf{x}}_0=(1/2,-1/2,1/2)$, direct evaluation givesWith ${\mathbf{s}}_0={\mathbf{x}}_1-{\mathbf{x}}_0$, the required linear system is thereforeAfter solving this linear system, form $\boxed{{\mathbf{x}}_1={\mathbf{x}}_0+{\mathbf{s}}_0}$. To get the second iterate, evaluate both the Jacobian matrix and the residual anew at ${\mathbf{x}}_1$, solveand set $\boxed{{\mathbf{x}}_2={\mathbf{x}}_1+{\mathbf{s}}_1}$. Reusing the initial Jacobian matrix would define a different iteration. The requested linear system has deliberately been left unsolved.