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

The second-derivative criterion for convexity isThe non-strict inequality is necessary and sufficient; the stronger condition $f^{\prime \prime }>0$ is sufficient for strict convexity.

Expand before taking the second derivative:The second-derivative criterion for convexity proves that $f$ is strictly convex, hence convex.

The Hessian criterion for convexity, in the usual twice-differentiable multivariate setting, isThus the Hessian matrix must be a positive semidefinite matrix everywhere. The criterion follows by applying the second-derivative criterion for convexity to each line restriction $t\mapsto g(x+tv)$, whose second derivative is $v^{\mathsf{T}}H(g)(x+tv)v$. Continuous second partials are a sufficient regularity hypothesis for this line-restriction argument.

Writing $g=e^{x^{\mathsf{T}}x}$, the gradient and Hessian matrix areConsequently, for any nonzero $v$,In particular the Hessian matrix is a positive semidefinite matrix everywhere, so the Hessian criterion for convexity proves $g$ is convex. The strict inequality along every nonconstant line also proves strict convexity.

The map $u\mapsto u^2$ is strictly increasing on $[0,\infty )$: for $u,v\ge 0$, $u\le v$ if and only if $u^2\le v^2$. Therefore a point has the smallest $f$-value if and only if it has the smallest $f^2$-value, and the same applies to $g$. By minimizers under strictly increasing transformations, both asserted argmin equalities hold, including when the argmin sets are empty. The differentiability assumptions are unnecessary for this order argument.

Minimize the square of the Euclidean distance, using minimizers under strictly increasing transformations. For $s=x+y>0$,For fixed $s$ the unique minimum has $x=y=s/2$. Put $x=y=\alpha >0$ and $J(\alpha )=2{\alpha }^2+{\ln }^2(2\alpha )$. Its derivative isLet $h(\alpha )=2{\alpha }^2+\ln (2\alpha )$. Since $h^{\prime }(\alpha )=4\alpha +1/\alpha >0$, it is strictly increasing, with limits $-\infty$ and $+\infty$ at the two ends of $(0,\infty )$. It therefore has exactly one zero. The sign of $J^{\prime }$ changes there from negative to positive, proving the unique global minimum. The nearest point is $P=(\alpha ,\alpha ,\ln (2\alpha ))$ at that zero.^[ref]

The Newton root-finding iteration isStarting at $0.4$ gives $0.376376476$, $0.376544572$, and $0.376544582$. Moreover $h(0.3765)<0<h(0.3766)$ certifies the rounding. Thus $\alpha =0.377$ to three significant figures; more accurately $P\approx (0.3765445825,0.3765445825,-0.2835716452)$.