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Because $k>0$ and the original smooth closed curve has periodic Frenet data, $\arctan (\tau /k)$ is a continuous real-valued periodic function. Therefore $\int k_{g}d\bar{s}=-\oint d\arctan (\tau /k)=0$. The simple closed normal curve divides the sphere into two disks by the Jordan theorem. Apply Gauss-Bonnet theorem to either disk, choosing its consistent boundary orientation. Its Gaussian curvature is one and its total boundary curvature remains zero. Thus each region has area $2\pi$, proving the two regions have equal area.

Let $T,n,B$ be the Frenet frame of the original space curve, with $T^{\prime }=kn$, $n^{\prime }=-kT+\tau B$, $B^{\prime }=-\tau n$. Put $w=\sqrt{k^2+{\tau }^2}$. The normal-indicatrix arc-length satisfies $d\bar{s}/ds=w>0$, and its unit tangent is $\overline{T}=(-kT+\tau B)/w$. On the unit sphere its outward normal is $n$, and $\overline{T}\times n=(-kB-\tau T)/w$.

Differentiate $\overline{T}$ in $s$. The derivative of $1/w$ is parallel to $\overline{T}$ and contributes nothing to its projection onto $\overline{T}\times n$. The remaining derivative is $(-k^{\prime }T+{\tau }^{\prime }B-(k^2+{\tau }^2)n)/w$, givingWith the alternative left-normal signed-curvature convention the displayed scalar changes sign, while the area conclusion is unchanged.

For an oriented surface normal $N$ and a unit-speed curve tangent $T$, take signed geodesic curvature $k_g=(dT/ds)\cdot (T\times N)$. This convention uses boundary orientation with the region on the right; reversing to the usual left-hand boundary convention reverses both orientation and the corresponding curvature convention. With positive orientation consistent this way, the Gauss-Bonnet theorem for a region $D$ with piecewise smooth boundary isFor a disk with smooth boundary there are no corner terms and $\chi (D)=1$. Stating the signed convention is necessary to reproduce the sign in part (ii).

Put $\zeta =e^{2\pi i/7}$ and induce the characters ${\chi }_r(a)={\zeta }^r$ of $H=⟨a⟩$. Since $H$ is normal, the induced value outside $H$ is zero, while at $a^m$ it is ${\zeta }^{rm}+{\zeta }^{2rm}+{\zeta }^{4rm}$. The nonzero $r$ split into the two doubling orbits, so the two distinct induced characters have class valueswhere $A=\zeta +{\zeta }^2+{\zeta }^4$ and $B={\zeta }^3+{\zeta }^5+{\zeta }^6=\overline{A}$. Summing seventh roots gives $A+B=-1$, and multiplication gives $AB=2$, hence $|A{|}^2=|B{|}^2=2$. Their character norms are $(9+3\cdot 2+3\cdot 2)/21=1$, proving irreducibility. Together with the three degree-one characters, the sum of squared degrees is $3+2\cdot 9=21$. Thus these are all the remaining irreducible characters.

For a character $\chi$ of a representation of $H\le K$, the induced character isThe sum is zero when no conjugate of $g$ meets $H$.

In a one-dimensional representation, the images commute, so $ba=a^{2}b$ forces the image of $a$ to be one. With $\omega =e^{2\pi i/3}$ the three possibilities areThey are inequivalent because they have distinct values at $b$.

Every group element has a unique form $a^r{b}^s$, with $r\mathrm{mod}7$ and $s\mathrm{mod}3$. Conjugation by $b$ doubles the exponent of $a$, giving the two orbits $\{1,2,4\}$ and $\{3,6,5\}$. Conjugation by powers of $a$ runs through all seven elements of each nonidentity coset of $⟨a⟩$. The classes are thereforeTheir sizes $1,3,3,7,7$ sum to twenty-one.

The energy equation rearranges to $p^2+(E+1)q^2=E{\mu }^2$, an ellipse with semiaxes $p_0=\mu \sqrt{E}$ and $q_0=\mu \sqrt{E/(E+1)}$. At low energy it is nearly circular with both radii $\mu \sqrt{E}$; at high energy it is tall and thin, with $q_0\to \mu$ while $p_0$ grows. The area enclosed is $\pi p_0{q}_0$, soConsequently $T\sim \pi {\mu }^2/(2\sqrt{E})=\boxed{\pi {\mu }^3/(2p_0)}$ at high energy.

The action variable for a regular periodic one-degree-of-freedom Hamiltonian orbit is $I(E)=(2\pi {)}^{-1}\oint pdq$, with orientation along the motion. Along its energy curve, implicit differentiation gives $\partial p/\partial E=1/H_p=1/\dot{q}$. Differentiating the loop integral branchwise givesAt simple turning points the two momentum branches join and their endpoint contributions cancel, so the integrable turning-point singularities do not spoil this differentiation. The formula concerns regular periodic levels, not a separatrix of infinite period.

For positive integers $n$, the finite open segment from part (i) works because $k=n/2>0$. For $n\le 0$, take a finite closed contour enclosing both branch points $i,-i$, with the integrand analytic in the exterior of the joining cut. Near infinity choose $(t^2+1{)}^{n/2-1}=t^{n-2}(1+t^{-2}{)}^{n/2-1}$. Since $n$ is integral, a circuit around both branch points returns to the initial branch, and the endpoint boundary term cancels. Thus this is an admissible finite contour, including $n=0$.

For $n\le -1$, at $z=0$ that exterior branch is $O(t^{n-2})=O(t^{-3})$. Deforming the contour to a large circle makes its integral tend to zero; equivalently its Laurent series has no $t^{-1}$ coefficient. Hence $\boxed{w(0)=0(n\le -1)}$. An individual loop around one branch point would not generally return to the same branch when $n$ is odd, which is why both are enclosed.

For $\mathrm{Re}k>0$, choose the straight path from $-i$ to $i$ and put $t=is$, $-1\le s\le 1$. Choose the power of $1-s^2$ continuously, positive for real positive $k$. The boundary factor vanishes at both endpoints. At zero,using the symmetric beta function integral. This choice also gives a nonzero solution.

With response rates and elasticity effectively fixed, changing ${\lambda }_2$ changes $m_2$ and hence intrinsic noise but not the environmental term. Observation (i) therefore indicates extrinsic noise dominates, making the mean-dependent intrinsic contribution negligible. For a general feedback law whose elasticity changes substantially with the mean, compensation between terms is another possibility; independence from ${\lambda }_2$ alone would then not logically identify the dominant term.

Observation (ii) measures the local relaxation rate ${\beta }_{2}H$, which is larger under negative feedback. Observation (iii) is not explained by faster adjustment alone. At fixed susceptibility, increasing the response rate increases ${\beta }_{2}H/({\beta }_1+{\beta }_{2}H)$, reducing time averaging and increasing transmitted environmental noise. Feedback lowers variance primarily by reducing susceptibility and the intrinsic term. Thus rapid adjustment and low noise share a feedback mechanism but are not a direct causal equivalence.

For the constitutive system, every propensity is affine in the copy numbers. The exact first- and second-moment equations therefore close: expectations of drift are determined by the means, and covariance drift by the covariance. The stationary fluctuation-dissipation calculation is consequently exact, even though the full joint stationary distribution need not be Gaussian.

For feedback, expectations such as $E[x_{1}f(x_2)]$ involve nonlinear moments and correlations. Replacing them by a linear expansion about the mean is a linear noise approximation, so the covariance formula is approximate. It works well around a stable single equilibrium when relative fluctuations are small and the rates vary smoothly over their typical range, generally at sufficiently large molecule numbers. Low counts, strong nonlinearities, multiple attractors and proximity to instability can invalidate it.

Factor the extrinsic term asThe first factor is environmental relative variance. The second is static susceptibility squared: differentiating the equilibrium relation gives $\delta \log m_2=H^{-1}\delta \log m_1$. The last is the time-averaging factor with effective protein response time $1/({\beta }_{2}H)$. Feedback makes response faster, which by itself weakens averaging, but simultaneously reduces the static gain. These are distinct effects and both must be retained.

At the deterministic feedback equilibrium use ${\lambda }_2{m}_{1}f(m_2)={\beta }_2{m}_2$. The indicated elasticity isThe normalized drift becomes $A=\left(\begin{array}{cc}-{\beta }_1& 0\\ {\beta }_2& -{\beta }_{2}H\end{array}\right)$; equal mean birth and death rates still give $D=\mathrm{diag}(2{\beta }_1/m_1,2{\beta }_2/m_2)$. Solving the same covariance equation yields the feedback attenuation of gene-expression noise:For $f(x_2)=k/x_2$, $h=1,H=2$, soThe inverse-feedback expression is interpreted near a positive equilibrium; a microscopic birth rate singular at zero needs a separate regularization there.

If RNA concentration does not fluctuate, protein synthesis is an immigration-death process, so its intrinsic normalized variance is $\boxed{1/m_2}$. If protein birth/death noise is suppressed and protein responds deterministically to the fluctuating RNA input, the remaining normalized variance isWriting ${\tau }_i=1/{\beta }_i$, its factor is ${\tau }_1/({\tau }_1+{\tau }_2)$. Slow environmental changes are followed almost fully; rapid environmental fluctuations are averaged over the longer protein response time. Thus extrinsic noise is the magnitude of relative RNA fluctuations multiplied by a dynamical averaging factor.

Let $m_i=⟨x_i⟩$. The mean equations give $m_1={\lambda }_1/{\beta }_1$ and $m_2={\lambda }_2{m}_1/{\beta }_2$. For relative deviations, the drift Jacobian and normalized reaction-noise covariance areThe normalized stationary fluctuation-dissipation relation for a reaction network is $A\eta +\eta A^T+D=0$ with this negative-drift convention. Its entries successively giveThe birth and death channels act on one species at a time, so the diffusion matrix is diagonal even though the stationary covariance is not.

For the radiation-era law beginning at $t_r$, the present particle-horizon estimate isRescale this present physical distance back to $t_r$ using $a(t_r)/a(t_0)=\sqrt{t_r/t_0}$. Its physical size was $2c\sqrt{t_r{t}_0}\simeq 0.6\mathrm{m}$, of order one metre as required. The matched inflationary Hubble rate is $H=1/(2t_r)\simeq 5\times 10^{34}{\mathrm{s}}^{-1}$, with Hubble radius $c/H\simeq 6\times 10^{-27}\mathrm{m}$.

If inflation produces $N$ e-folds before $t_r$, this patch began with size $0.6e^{-N}$ metres. It fits within one Hubble radius ifThis estimate uses the idealized radiation-only history stated in the question; substituting the actual matter-dominated history would change the numerical estimate.

For an expanding decelerating power law $a\propto t^{\alpha }$ with $0<\alpha <1$, the particle horizon is $ct/(1-\alpha )$, and the comoving Hubble radius is $c/(aH)\propto t^{1-\alpha }$. It increases with time, so regions that are mutually visible today can have been out of causal contact at early times. The nearly uniform early thermal conditions inferred across such regions give the horizon problem.

During cosmic inflation, accelerated expansion makes the comoving Hubble radius decrease. A patch initially within one causal communication scale can expand into the region now observed, after which its separated parts are temporarily outside the Hubble radius and later re-enter during deceleration. Sufficient duration before the hot radiation era permits a common causal origin; accelerated expansion is not itself a claim that signals propagate faster than light locally.

For a homogeneous isotropic metric, radial light propagation covers comoving distance $\int cdt/a(t)$. The physical particle horizon is $d_p(t)=a(t){\int }_{t_i}^{t}cd{t}^{\prime }/a(t^{\prime })$, when this is finite. A future event horizon exists when $d_e(t)=a(t){\int }_t^{t_f}cd{t}^{\prime }/a(t^{\prime })$ is finite. The Hubble radius is the instantaneous scale $c/H$, with $H=\dot{a}/a$; it is not generally the same as either causal horizon.

For an Einstein-de Sitter universe, $H=2/(3t)$, so $c/H=3ct/2$, while $d_p=3ct$ from a beginning at zero. The future integral diverges, so there is no event horizon. For an eternal-past de Sitter spacetime, $H$ is constant and the particle-horizon integral from $-\infty$ diverges. Its event horizon and Hubble radius both equal $c/H$. Thus a Hubble radius is not automatically a particle horizon, even in these simple isotropic models.