The derivative isA stationary point at $x=1$ requires $2/a^2-2a^2=0$, or $a^4=1$. Since $a>0$, $\boxed{a=1}$.

Since $\mathbf{v}=-\mathbf{u}$ and $\mathbf{u}\ne 0$, the vectors point in opposite directions. More explicitly, their dot product givesHence their angle is $\boxed{\theta =\pi \text{ radians}=180^{\circ }}$.

Adding the vectors component by component givesConsequently $\boxed{a=0}$. The requirement that the vectors be nonzero then implies $b\ne 0$.

For $a=b=1$, the stationary points found in parts (a) and (b) have temperatures $0$, $1$, $1/4$ and $1/5$. The unit sphere is compact and $T$ is a continuous function, so the extreme value theorem ensures that a hottest point exists; it must be one of those stationary points. ThusFor a direct bound, put $s=x^2\in [0,1]$. The identity $(y^2-z^2{)}^2\ge 0$ gives $y^2{z}^2\le (y^2+z^2{)}^2/4=(1-s{)}^2/4$, andsince $1-s^2-(1-s{)}^2/4=(1-s)(5s+3)/4\ge 0$. Equality forces $s=1$, giving exactly the two points above.

Solve the Lagrange multiplier equations by separating cases according to which coordinates vanish, so that division by a coordinate loses no stationary points.

If $x\ne 0$ and $y=z=0$, the surface constraint givesIf $x\ne 0$ and exactly one of $y,z$ is nonzero, the equation for that nonzero coordinate forces $\lambda =0$, while the $x$ equation gives $\lambda =2a^2{x}^2>0$. Hence there are no stationary points in that case.

If $x=0$ and $y,z\ne 0$, the two remaining equations give $\lambda =a^2{z}^2=b^2{y}^2$. The surface constraint therefore has equal contributions from $y$ and $z$, giving four stationary points:If all three coordinates are nonzero, the equations implyso $z^2=2x^2$ and $y^2=2a^2{x}^2/b^2$. Substitution in the surface constraint gives, with $D=4a^2+b^2$,Thus the remaining eight stationary points and their temperatures areFor example, their common temperature follows from $x^4+y^2{z}^2=a^4{b}^4/D^2+4a^6{b}^2/D^2=a^4{b}^2/D$. The cases with $x=0$ and $yz=0$ are exactly the coldest points already listed. This exhausts all possibilities, giving fourteen other stationary points in addition to those four.

Let $G=x^2/a^2+y^2/a^2+z^2/b^2-1$. Its gradient cannot vanish on the ellipsoid, since the origin is not on the surface. Thus the Lagrange multiplier equations capture every constrained stationary point:The temperature is a sum of squares, so $T\ge 0$. Equality requires $x=0$ and $yz=0$. The surface constraint then gives the four coldest points:These are all such points, because a sum of nonnegative terms can vanish only when each term vanishes.

By part (e) and the comparison test for series, $\sum u_n$ converges, since the P-series $\sum n^{-2}$ converges. Put $S={\sum }_{n=1}^{\infty }{u}_n$. Part (a) now provesPart (d) also shows that $(a_n)$ increases towards this finite positive limit. This proves convergence of the Stirling sequence without presupposing its limiting constant.

For identification of that constant, the independently known Stirling's formula $n!\sim \sqrt{2\pi n}(n/e{)}^n$ givesThe comparison argument above supplies the convergence proof requested by the preceding parts; the Stirling's formula identification is an additional evaluation.

For $k\ge 2$, writeUse positivity from part (d) and bound the absolute terms of the power series in part (c) by a geometric series. For $n\ge 2$,For $n=1$, $u_1=\frac{3}{2}\ln 2-1<\frac{1}{2}<1$, since $\ln 2={\int }_1^{2}dt/t<1$. Thus a bound valid for every positive integer $n$ is

A rational lower bound for the natural logarithm proves positivity for every $n$, including $n=1$. DefineThen $h(0)=0$, while direct differentiation givesThus $h(t)>0$ for $t>0$. Setting $t=1/n$ in the expression from part (b) givesTherefore $\boxed{u_n>0\text{ for every }n\ge 1}$, and $(a_n)$ is a strictly increasing sequence.

For $n>1$, the Taylor series of the natural logarithm converges absolutely at $1/n$:Multiply this power series by $n+1/2$ and subtract $1$. The constant term cancels the subtracted $1$, and the coefficient of $1/n$ is $-1/2+1/2=0$. For $k\ge 2$, the coefficient of $n^{-k}$ isHenceIn particular $u_n=1/(12n^2)-1/(12n^3)+O(n^{-4})$, in Big O notation.

Cancellation of the factorials in the consecutive ratio givesSet $t=1/n$. The derivative of the natural logarithm at $1$ gives $\ln (1+t)/t\to 1$. ThereforeThus the necessary condition from the term test for divergence is satisfied. On its own it does not prove that $\sum u_n$ converges.

All $a_n$ are positive, and $a_1=e^{-1}$. The logarithm turns consecutive ratios into differences, producing the finite telescoping sumThus, already for every finite $n$,If the partial sums converge to $S$, the continuity of the exponential function givesThe shift from $a_{n+1}$ to $a_n$ preserves the limit of a sequence. Convergence of the partial sums is established in part (f), so no convergence assumption is left unsupported.

At $x=\pi /2$, all odd-indexed cosine terms vanish, while $\cos (2k\pi /2)=(-1{)}^k$. Thus the Fourier series from part (c) givesDivide by $2(e^{\pi }-1)/\pi$ and rearrange. The hyperbolic sine identity $e^{\pi }-1=2e^{\pi /2}\sinh (\pi /2)$ yieldsThe evaluation is legitimate by the convergence established in part (c).

The function is even, so each product with a sine is an odd function and all the Fourier coefficients $b_n$ vanish. Its constant coefficient isFor $n\ge 1$, part (a) gives the remaining Fourier coefficients:Consequently the Fourier cosine series isThe periodic extension is continuous, including at $\pm \pi$, and piecewise continuously differentiable. Thus the Fourier series equals $f$ everywhere. Its coefficients are also bounded in magnitude by a constant times $1/n^2$, which gives absolute convergence.

A Fourier series represents a periodic function as a constant plus sine and cosine waves whose frequencies are integer multiples of its fundamental frequency. For period $2L$, use the conventionwhere the Fourier coefficients areThe coefficient formulas follow by multiplying by a basis sine or cosine and using their orthogonality over a full period. A Fourier series need not equal an arbitrary function at every point. The pointwise convergence of a piecewise smooth Fourier series theorem says that for a piecewise continuously differentiable function, the Fourier series converges at each point to the average of the two one-sided limits, and hence to $f(x)$ wherever $f$ is continuous.

From part (f), $PQ=\mathrm{diag}(-1,-1,1)$. To obtain $PQR=-I$, choose the Householder transformation that reverses only the third coordinate:Indeed $\mathbf{r}{\mathbf{r}}^T=\mathrm{diag}(0,0,2)$, so $R_{ij}={\delta }_{ij}-r_i{r}_j$, and direct matrix multiplication gives $PQR=-I$. The alternative vector $-\mathbf{r}$ produces the same matrix. Geometrically $R$ reflects in the plane $x_3=0$, completing the reversal of all three coordinates.

The second matrix acts byApplying $P$ from part (e) then givesThe Householder transformation $Q$ reflects in the plane $x_1-x_2=0$. Its normal vector $\mathbf{q}$ is perpendicular to $\mathbf{p}$, so the two reflecting planes meet at a right angle along the $x_3$ axis. Their composition is a rotation through $\pi$ about the $x_3$ axis, as is also immediate from the displayed rotation matrix.