Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics and probability resit / 4

Let $e_n=x_n-x_{\ast }$ for the limit $x_{\ast }$. Quadratic convergence means that eventually $|e_{n+1}|\le C|e_n{|}^2$ for a fixed constant $C$; exact asymptotic order two has $|e_{n+1}|/|e_n{|}^2\to c$ with $0<c<\infty$.

For this iteration and $x_{\ast }=3$, direct algebra givesUsing the stipulated convergence $x_n\to 3$, the ratio of successive squared errors isThus the convergence is quadratic. The iteration is well defined here: $x_1=7/2$, and thereafter $x_n>3$ implies $3<x_{n+1}<x_n$. The initial point is not already the limit, so the error ratio is meaningful.

Assume $f$ and $g$ are continuously differentiable near the point. At a regular feasible point, meaning $g(x)=0$ and $\nabla g(x)\ne 0$, a differentiable constrained local extremum satisfies the Lagrange multiplier equationsThey arise by differentiating the optimization Lagrangian $L(x,\lambda )=f(x)-\lambda g(x)$ with respect to all coordinates and the multiplier. Geometrically, the objective has zero derivative along the tangent space of the constraint surface, so its gradient must lie in the normal direction.

Solve these equations for candidates, examine feasible points where the regularity condition fails separately, and compare objective values or use a second-order test restricted to feasible directions. The equations locate stationary points; they do not by themselves establish which are minima or whether a global minimum is attained.

Use $F=x^6+\frac{3}{2}{y}^4$ and $G=x^4-2y^2-3$. Every feasible point has $x\ne 0$, so $\nabla G=(4x^3,-4y)$ cannot vanish. The Lagrange multiplier equations areThe first gives $\lambda =\frac{3}{2}{x}^2>0$. If $y\ne 0$, the second would give $\lambda =-\frac{3}{2}{y}^2<0$, a contradiction. Therefore $y=0$, and the constraint gives $x^4=3$. The only constrained stationary points are $(\pm 3^{1/4},0)$.

For a global certificate, feasibility implies $x^4\ge 3$, and henceEquality requires $y=0$ and $x^4=3$. ThusThe case split is essential: dividing by $y$ would discard precisely the minimizers.

Here the constraint surface $x^4+2y^2=3$ is compact and regular, so a global minimum exists and satisfies the Lagrange multiplier equationsKeep the zero-coordinate cases. If $y=0$, the two points are $(\pm 3^{1/4},0)$, with value $3\sqrt{3}$. If $x=0$, the two points are $(0,\pm \sqrt{3/2})$, with value $27/8$. If both coordinates are nonzero, the equations give $x^2=y^2=t>0$. The constraint becomes $t^2+2t=3$, whose only positive solution is $t=1$. These are the four points $(\pm 1,\pm 1)$, all with value $5/2$.

These cases exhaust the stated eight constrained stationary points. Comparing their values givesFor additional classification, substituting $u=x^2\in [0,\sqrt{3}]$ gives $F(u)=u^3+\frac{3}{8}(3-u^2{)}^2$, withIt decreases from $u=0$ to $u=1$ and increases from $u=1$ to $u=\sqrt{3}$. Thus the two $y$-axis points are constrained local maxima, the two $x$-axis points are global maxima, and the four mixed-coordinate points are global minima.

Both terms $x^6$ and $\frac{3}{2}{y}^4$ are nonnegative, and they vanish simultaneously only at $(0,0)$. That point satisfies $x^4+2y^2\le 3$. Consequentlyand the global minimum is unique. No Lagrange multiplier calculation is needed: nonnegativity gives a lower bound attained at a feasible point.