= Solution
{c}
Expanding and taking <partial derivatives> gives the <gradient> and <Hessian matrix>
$$
\nabla f=\begin{pmatrix}2x+2y^2+4x^3+4xy\\4xy+4y^3+2x^2+2y\end{pmatrix},
\qquad
H=\begin{pmatrix}2+12x^2+4y&4x+4y\\4x+4y&2+12y^2+4x\end{pmatrix}.
$$
At $p_0=(-1,0)^T$, these are $\nabla f(p_0)=(-6,2)^T$ and
$$
H(p_0)=\begin{pmatrix}14&-4\\-4&-2\end{pmatrix},\qquad \det H(p_0)=-44\ne0.
$$
Thus the <Newton step> solves
$$
\begin{pmatrix}14&-4\\-4&-2\end{pmatrix}
\begin{pmatrix}s_x\\s_y\end{pmatrix}
=\begin{pmatrix}6\\-2\end{pmatrix}.
$$
The second equation gives $s_y=1-2s_x$; substituting into the first gives $22s_x=10$. Hence
$$
s_x=\frac5{11},\qquad s_y=\frac1{11},\qquad
\boxed{(x_1,y_1)=\left(-\frac6{11},\frac1{11}\right).}
$$
At $p_*=(-1/2,-1/2)^T$, however,
$$
\nabla f(p_*)=0,\qquad
H(p_*)=\begin{pmatrix}3&-4\\-4&3\end{pmatrix}.
$$
The <eigenvalues> of this <Hessian matrix> are $-1$ in direction $(1,1)$ and $7$ in direction $(1,-1)$. It is invertible but indefinite. Thus the <Newton step> is zero, and
$$
\boxed{(x_n,y_n)=\left(-\frac12,-\frac12\right)\quad\text{for every }n.}
$$
This illustrates that <stationary points are fixed points of Newton's method>. Here the fixed point is a \b[<saddle point of a scalar function>], not a <local minimum>. Indeed, direct substitution along the two <eigenvector> directions gives
$$
\begin{aligned}
f(-\tfrac12+t,-\tfrac12+t)&=\frac18-t^2+2t^4,\\
f(-\tfrac12+t,-\tfrac12-t)&=\frac18+7t^2+2t^4.
\end{aligned}
$$
For sufficiently small nonzero $t$, one direction decreases $f$ and the other increases it. The <global minimum> is instead zero: the sum of squares vanishes at $(0,0)$ and $(-1,-1)$. These are the only real zeros, since $x=-y^2$ and $y=-x^2$ imply $y(y^3+1)=0$. Thus a zero <gradient> is an insufficient stopping certificate for <minimization problem> without classifying the <stationary point>.
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