Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 4

One standard finite form of Taylor's theorem is the following. Suppose $f$ has continuous derivatives through order $m+1$ on an interval containing $a$ and $x$. Then, if $x\ne a$, there is some $\xi$ strictly between them such thatThe final term is the Lagrange remainder. At $x=a$ the identity holds directly with zero remainder. The finite expansion requires no assumption that an infinite Taylor series converges to $f$.

For $0<x<\pi$, apply Taylor's theorem to the sine about zero, retaining terms through degree four. Its fourth-order coefficient is zero, and its fifth derivative is the cosine function, givingThe cosine function is decreasing on $[0,\pi ]$, so $\cos x\le \cos \xi \le 1$. Multiplying by the nonnegative factor $x^5/120$ and adding $x-x^3/6$ proves both required bounds. At $x=0$ both are equalities.

Let $h(x,y)=2y^3+x{y}^2-y^2-4x$, so $g=e^h>0$. Its gradient is $\nabla g=g\nabla h$, andThus a stationary point must have $y=\pm 2$ and $x=1-3y$, giving $(-5,2)$ and $(7,-2)$. To classify them, useAt either stationary point, $\nabla h=0$ and $12y+2x-2=6y$. Hence $\det H_g=g^2(-4y^2)=-16g^2<0$. The Hessian has one positive and one negative eigenvalue, so both are saddle points. There are no other stationary points.

Let $q=x^2+4y^2$. The objective has the exact sum of squares decompositionFor every feasible point, $q\le 8$, so this is at least $3-8/2=-1$. Equality requires simultaneously $q=8$ and $x+2y=0$. Substitution gives $y^2=1$ and $x=-2y$. Both resulting points are feasible. Therefore the global minimum is $-1$, attained exactly at $(2,-1)$ and $(-2,1)$. The lower bound and its equality conditions prove global attainment of the minimum, rather than merely identifying stationary points on the boundary.