The matrix issoHere $\Vert \mathbf{p}{\Vert }^2=2$, so $P=I-2\hat{\mathbf{p}}{\hat{\mathbf{p}}}^T$ with $\hat{\mathbf{p}}=\mathbf{p}/\sqrt{2}$. This is a Householder transformation: reflection in the plane $x_1+x_2=0$. The component parallel to the normal vector $\mathbf{p}$ changes sign, while the perpendicular component is fixed.

Using Einstein notation and the Kronecker delta,The unit vector condition is essential in the last step.

Apply matrix multiplication to obtainSince $\hat{\mathbf{n}}$ is a unit vector, the subtracted term is the orthogonal projection of $\mathbf{x}$ along $\hat{\mathbf{n}}$. Hence $M$ is the orthogonal projection onto a plane through the origin with normal vector $\hat{\mathbf{n}}$. It fixes every vector perpendicular to $\hat{\mathbf{n}}$ and sends every scalar multiple of $\hat{\mathbf{n}}$ to zero. In particular $\hat{\mathbf{n}}\cdot M\mathbf{x}=0$.

By the definition of matrix multiplication and trace, with all indices summed from $1$ to $D$,The entries are scalars, so their multiplication commutes. Reordering them and renaming the summed indices givesEquivalently, in Einstein notation, $P_{ij}{Q}_{jk}{R}_{ki}=Q_{jk}{R}_{ki}{P}_{ij}$. This proves the cyclic property of the trace.

The gradient isIt is perpendicular to each nonzero level set and points towards increasing $f$. Near the two positive local maxima, the Hessian matrix is $-2e^{-1}I$, so the gradient arrows point inward. Near the two negative local minima, the Hessian matrix is $2e^{-1}I$, so they point outward. These arrows are shown on the figure in part (c).

At the saddle point, $\nabla f\approx (v,u{)}^T$. Along $v=u$, the arrows point away from the centre on both sides; along $v=-u$, they point towards the centre on both sides. On the positive $u$ axis they point up, on the negative $u$ axis down, on the positive $v$ axis right, and on the negative $v$ axis left. This specifies the four intervening sectors of the gradient pattern.

In the translated coordinates $u=x-1$, $v=y+1$, the zero level set is the pair of lines $u=0$ and $v=0$. The function is positive in the quadrants $uv>0$ and negative in those with $uv<0$, and tends to zero as $u^2+v^2\to \infty$. Thus its nonzero contours form closed curves around the four off-centre stationary points.

For an explicit description of the contour plot, use polar coordinates $u=r\cos \theta$, $v=r\sin \theta$:At fixed direction within a quadrant, the magnitude increases until $r=\sqrt{2}$ and then decreases towards zero. Also $|u|e^{-u^2/2}\le e^{-1/2}$, with equality exactly when $|u|=1$, and similarly for $v$. Consequently $|f|\le e^{-1}$; each positive level below $e^{-1}$ encloses a positive peak in each positive quadrant, with the analogous negative contours in the negative quadrants.

To classify the stationary points, the Hessian matrix isAt $u=v=0$ it is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with eigenvalues $1$ and $-1$, so $(1,-1)$ is a saddle point. At $u,v\in \{-1,1\}$ it is $-2fI$. Therefore $(2,0)$ and $(0,-2)$ are strict local maxima, and $(2,-2)$ and $(0,0)$ are strict local minima. These are also global maxima and minima by the bound just proved.

At the central stationary point, $u=v=0$, so $f=0$. At each of the other stationary points, $u^2+v^2=2$ and hence $f=uv/e$. Therefore

Translate the coordinates by $u=x-1$ and $v=y+1$, and write $E=e^{-(u^2+v^2)/2}>0$. The partial derivatives areIf $u=0$ or $v=0$, these equations together force $u=v=0$. Otherwise they require $u^2=v^2=1$. Thus the complete set of stationary points is

For the given differential form, set $P=x(y^4+1)$ and $Q=y(2x^2{y}^2+1)$. Their partial derivatives satisfyIn fact an explicit potential issince $F_x=P$ and $F_y=Q$. Thus the differential form is an exact differential form on all of ${\mathbb{R}}^2$.

Replacing $y^2$ by $u$ suggests the potentialDirect differentiation verifies that $dH/dx$ is exactly the left-hand side of the required ordinary differential equation. This direct verification remains valid even for negative $u$; it does not require a real square root $y=\sqrt{u}$. ThereforeThe initial condition gives $C=1$. Solving this quadratic equation in $u$ and selecting the branch that has $u(1)=0$ givesThe second expression also defines the removable value $u(0)=1$. The maximal open interval containing $1$ on which this is a differentiable real solution isAt either endpoint the square root vanishes and $2x^{2}u+1=0$, while $x(u^2+1)\ne 0$, so the original ordinary differential equation cannot hold there with a finite derivative.

Keep $a$ general initially, since the final part does not explicitly fix it. For $-h\le y\le h$, where $h=\pi /6$, the cumulative distribution function isThe normalization constant in the probability density function of $Y$ must beEquivalently, the integrated cumulative distribution function on a bounded interval identity gives $M=h-\mathbb{E}[X]$. If the assumption $a=0$ from part (c) is retained, the requested special case is $\boxed{M=\pi /6}$. For the value $a=1$ in part (b), it is $M=\pi /6-{\pi }^3/324$.

Here $a=0$, so the probability density function is the even function $\cos x$ on $[-h,h]$, where $h=\pi /6$. Hence $\mathbb{E}[X]=0$ by oddness, and the variance is just $\mathbb{E}[X^2]$. Two applications of integration by parts giveThus

For $h=\pi /6$, the expected value isbecause $x\cos x$ is an odd function. Comparing with the specified expected value gives $\boxed{a=1}$, which lies in the permitted parameter range.

Use the tangent half-angle substitution $t=\tan (x/2)$, so $\cos x=(1-t^2)/(1+t^2)$ and $dx=2dt/(1+t^2)$. ThenThe inverse tangent antiderivative givesThis expression applies on each interval $((2j-1)\pi ,(2j+1)\pi )$ for an integer $j$. Although $5+3\cos x\ge 2$ everywhere, the chosen inverse tangent expression jumps at the endpoints because the substitution has a pole there. A global continuous antiderivative is obtained by adding $j\pi /2$ on that interval, using one common $C$, and extending by continuity at the endpoints.

Split the integral intoThe first is the standard antiderivative $\mathrm{arsinh}x=\ln (x+\sqrt{1+x^2})$ from the inverse hyperbolic functions. For the second, the substitution rule $t=1+x^2$ gives $\frac{1}{2}\int t^{-1/2}dt=\sqrt{1+x^2}$. HenceThe logarithm's argument is positive for every real $x$.