Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 3

The Taylor expansions $\sin t=t+O(t^3)$ and $e^t=1+t+t^2/2+O(t^3)$ giveHence the second-order Taylor polynomial isThe sine contributes no quadratic term at zero.

For a twice continuously differentiable function, an interior local extremum requires $f_x=f_y=0$. At such a critical point, put $D=f_{xx}{f}_{yy}-f_{xy}^2$. The second-derivative test gives a strict local minimum if $D>0$ and $f_{xx}>0$, a strict local maximum if $D>0$ and $f_{xx}<0$, and a saddle point if $D<0$. When $D=0$, the test is inconclusive. More generally the Hessian matrix is necessarily positive semidefinite at a local minimum and negative semidefinite at a local maximum, but these non-strict signs alone are not sufficient.

For this polynomial,Subtracting the two stationarity equations gives $y(2x-y)=0$. If $y=0$, then $x=0$; if $y=2x$, the remaining equation gives $x=0$ or $x=3/2$. Thus the only critical points are $(0,0)$ and $(3/2,3)$.

At $(3/2,3)$, $\det H=-18<0$, so it is a saddle point. At $(0,0)$ the determinant is zero, but along $y=-x$ the value is $-x^3$, which takes both signs arbitrarily near zero. ThereforeIn particular a vanishing Hessian determinant does not exclude a saddle point.

Assume $f,F$ are continuously differentiable near the point and $F_z\ne 0$. The implicit function theorem supplies a local function $z(x,y)$ withAt an interior extremum of $g(x,y)=f(x,y,z(x,y))$, the multivariable chain rule givesMultiply these identities by $F_y$ and $F_x$, respectively, and subtract. The terms containing $f_z$ cancel, givingNo assumption $f_z\ne 0$ is needed for this elimination; dividing by $f_z$ would unnecessarily exclude valid cases. If $F_z=0$, this particular local graph argument is unavailable and a different chart or the full Lagrange multiplier condition is needed.

On the constraint $x^2-z^2=1$, substitute $x^2=1+z^2$ to obtainTherefore the two global minima areEvery pair $(y,z)$ is allowed on either smooth branch $x=\pm \sqrt{1+z^2}$, so these are also the only local extrema. There is no local maximum or global maximum: the displayed positive quadratic in $(y,z)$ has no other stationary point and is unbounded above. At the minima $F_z=-2z=1$, consistent with the hypothesis in part i.