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Put $c=\cos \theta$ and $s=\sin \theta$. Taking the real part of $(c+is{)}^6$ in De Moivre's theorem givesTaking the imaginary parts of the second and fourth powers gives $\sin 2\theta =2cs$ and $\sin 4\theta =4cs(c^2-s^2)$. HenceThe mixed powers cancel exactly.

Apply scalar multiplication and vector addition componentwise:

The Euclidean norm of the vector is the square root of the sum of its squared components:

Use the outward unit normal on the unit sphere and the convention that $\theta$ is measured from the positive vertical axis. Then $\mathbf{F}\cdot \mathbf{n}=\sin \theta$, while the spherical coordinates area element is $dS=\sin \theta d\theta d\varphi$. The flux is therefore

The vector field is radial, while the differential displacement along a centred circle is tangential. These directions are orthogonal, so the dot product vanishes at every point. Hence

Near the initial condition $y=1$, divide by $y^2$ and separate variables: $y^{-2}dy=-x^{2}dx$. Integration gives $-1/y=-x^3/3+C$; the initial value fixes $C=-1$. Thus the solution of the separable differential equation isThe stated interval is the maximal connected interval containing zero: the solution diverges at its left endpoint. The constant solution $y=0$ does not satisfy the initial condition.

The integrating factor for this linear ordinary differential equation is $e^x$. Multiplication gives $(e^{x}y{)}^{\prime }=1$, so $e^{x}y=x+C$. The initial condition sets $C=0$, giving

The vector field has only a vertical component, namely $f(r)$. Its divergence is therefore its vertical partial derivative:

For $r>0$, the partial derivatives of radius are $r_x=x/r$, $r_y=y/r$ and $r_z=z/r$. Applying the chain rule to the outer sine and then the radial profile gives the Cartesian coordinates of the gradient:At the origin these coordinate formulas require a differentiable radial extension; differentiability of $f$ as a one-variable profile alone does not guarantee it.

On a smooth implicit branch, put $Q=x+z^2$ and require $Q\ne 0$. This is exactly the implicit function theorem condition, since the defining equation has vertical partial derivative $3Q$. Differentiating the cubic relation in $x$ and $y$ givesFor a second derivative, the dependence of $Q$ on $z$ must also be included: $Q_x=1+2z{z}_x$ and $Q_y=2z{z}_y$. ThusThe required cancellation follows immediately:The verification is local to smooth branches with $x+z^2\ne 0$; the implicit relation does not guarantee a differentiable function at its singular points.

Write ${ϵ}_k=E_0(k+\frac{1}{2})$ and use Lagrange multipliers $\alpha ,\beta$ for the particle-number and energy constraints. The stationary equations areThus $n_k=e^{-\alpha }{e}^{-\beta {ϵ}_k}$. Finite normalization over infinitely many levels requires $\beta >0$. Put $q=e^{-\beta E_0}$, so $0<q<1$. The geometric series givesThereforeFor the energy constraint, use $n_k=N(1-q)q^k$ and $\sum k{q}^k=q/(1-q{)}^2$, obtained by differentiating the geometric series. This yieldsRecognizing the hyperbolic cotangent givesThe strict positivity of all occupations requires $E>N{E}_0/2$; the paper's positivity assumptions alone do not guarantee this feasibility condition. For admissible energy the multiplier is uniquely $\beta =E_0^{-1}\ln [(2E+N{E}_0)/(2E-N{E}_0)]>0$. Equality $E=N{E}_0/2$ is only the limiting state with all occupation in level zero, outside the stipulated strictly positive family; lower energy is impossible.

For rigor beyond formal multiplier equations, $p_k=n_k/N$ has fixed mean $E/(N{E}_0)-1/2$. The objective differs from $N$ times its information entropy only by the fixed constant $N-N\ln N$. The maximum entropy distribution on the nonnegative integers is uniquely geometric at that mean, confirming the stationary family and its maximum character.

At a fixed radius, Parseval's identity applied to the preceding Fourier series givesIn polar coordinates, $dA=rdrd\theta$. Integrating over radius and interchanging the nonnegative sum and integral givesThe interchange follows from the monotone convergence theorem. This is the special case of the area norm of the Poisson extension in the unit disk for the odd step boundary data.

The boundary data are odd except at isolated endpoint choices, which do not change integrals. Thus every cosine Fourier coefficient, including $a_0$, vanishes. For the sine Fourier coefficients,Consequently the extension isIt converges absolutely for every $r<1$; its limiting boundary values at jumps are the midpoint of the two one-sided limits.

For fixed $r<1$, $\sum r^n\cos (n\theta )$ converges absolutely and uniformly. Thus for integrable $f$, one may integrate the kernel series term by term. The cosine addition formula givesSubstitute this into the defining integral. The constant kernel term gives $\frac{1}{2\pi }\int f=a_0/2$; the remaining terms givewhereThese are the same Fourier coefficients as in part (a). The kernel $2K$ is the Poisson kernel for the unit disk; the factors $r^n$ suppress the higher angular frequencies inside the disk.

With the real Fourier series convention,where the Fourier coefficients areFor a square-integrable periodic function, Parseval's identity isThe Fourier coefficients require integrability, and the stated Parseval identity requires square integrability. The expansion agrees with $f$ in the L2 space sense; for piecewise smooth data it converges pointwise to the average of the two one-sided limits. Without such qualifications an arbitrary periodic function need not have a pointwise-convergent Fourier series.

At $\mu =-1$, ${\mathbf{c}}_2=-{\mathbf{c}}_1$, while ${\mathbf{c}}_1$ and ${\mathbf{c}}_3$ are independent. Their span is a plane. A solution requires $\mathbf{b}$ to lie in that plane, so every scalar triple product formed from two columns and $\mathbf{b}$ must vanish. The product with ${\mathbf{c}}_1,{\mathbf{c}}_2$ vanishes identically; the other two give the plane condition.

Explicitly, ${\mathbf{c}}_1\times {\mathbf{c}}_3=(-7,1,3)$, soThus $\lambda =7$ is necessary and, since the two columns span the plane, sufficient. The equations then give $x-y=1$ and $z=2$, soThe solution is an affine line parallel to the homogeneous null space from part (c), obtained by intersecting the compatible constraint planes. Every point on it is nonzero because $z=2$.