Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Discrete mathematics, continuous mathematics and probability / 6

Differentiating the polynomial coordinate map givesIts Jacobian matrix isThe determinant identifies where the inverse function theorem applies.

On a branch with fixed signs of $u,v$ and $uv\ne 0$, use $u^2=x+y$ and $v^2=y-x$. Differentiating these identities, or inverting the Jacobian matrix, givesThe inverse Jacobian matrix is thereforeThese are local partial derivatives. Choosing the signs specifies an inverse function on $y>|x|$; the original map has four preimages there and has no single global inverse. At $u=0$ or $v=0$, the displayed inverse derivatives are not finite.

The requested partial derivatives areThe last printed derivative has a misplaced superscript on $f$; the second-derivative list and subsequent Taylor expansion use $f_{yy}$.

Put $h=x-1$ and $k=y-1$. At $(1,1)$, the value is $5+4\alpha$, the gradient is $(5+4\alpha ,4+2\alpha )$, and the Hessian matrix is $\left(\begin{array}{cc}8& 2\alpha \\ 2\alpha & 2\end{array}\right)$. Thus the quadratic Taylor expansion isThe exact remainder is $h^3$, since the original polynomial has only the cubic term $x^3$ beyond degree two.

Solving $f_y=0$ gives $y=-1-\alpha x$. Substitution into $f_x=0$ gives $x\{3x+2(1-{\alpha }^2)\}=0$. Therefore $(0,-1)$ is always a critical point. When $\alpha \ne -1$ as well as the printed restriction $\alpha \ne 1$, there is a second critical pointAt $(0,-1)$, the Hessian matrix is $\left(\begin{array}{cc}2& 2\alpha \\ 2\alpha & 2\end{array}\right)$ with determinant $4(1-{\alpha }^2)$. The second-derivative test gives a strict local minimum when $|\alpha |<1$ and a saddle point of a scalar function when $|\alpha |>1$.

The allowed value $\alpha =-1$ requires higher-order terms. With $v=y+1+\alpha x$, completing the square gives the exact identityThe coordinate change is invertible, so the Hessian at a critical point under a change of coordinates preserves the classification. At $\alpha =-1$, this is a cubic degenerate saddle: along $v=0$, the difference from the critical value is $x^3$ and takes both signs. Thus, throughout the printed parameter range,There is no local maximum. When the second point is distinct, it is a local minimum for $|\alpha |>1$ and a saddle point of a scalar function for $|\alpha |<1$: in the $(x,v)$ coordinates its Hessian matrix is $\mathrm{diag}(-2(1-{\alpha }^2),2)$. The printed instruction to split only at $\alpha =1$ misses the negative-parameter cases.