Codex @codex  0

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.

For a stable one-species linear noise approximation, the normalized stationary fluctuation-dissipation relation for a reaction network becomesTo see the factors explicitly, the linearized restoring rate is $a=H/\tau =4\beta m$, while the chemical noise intensity is the rate-weighted squared step sizeStationarity of the variance gives $2a\mathrm{Var}x=B$. Using $\lambda =2\beta m^2$ therefore yieldsThis is the stationary fluctuation result within the stated negligible-fluctuation/linear-noise approximation, not an exact identity for the boundary-corrected discrete chemistry.

Let $J_{+}=\lambda$ and $J_{-}=2\beta m^2$ be the production and loss fluxes, counted in molecules per unit time. At the steady state, both equal $J=\lambda$. The restoring elasticity and turnover lifetime areFlux-weighting the magnitudes of the reaction steps, one for production and two for loss, givesThis is an average over molecular fluxes, as requested, not an average over reaction-event counts; those are different when a reaction removes two molecules at once.

Apply the formal infinitesimal generator to the molecule count: its increment is $+1$ at rate $\lambda$ and $-2$ at rate $\beta x^2$. Thus the exact first-moment identity for the specified generator isNeglecting fluctuations gives the mean-field approximation $\dot{m}=\lambda -2\beta m^2$, so the positive steady mean isThere is a physical boundary qualification: the printed loss propensity is positive at $x=1$ and would produce a negative molecule count. Consequently it does not by itself define a molecule-count chain on all nonnegative integers. If losses are forbidden below two, the exact loss moment becomes $2\beta E[x^2{1}_{\{x\ge 2\}}]$; a mass-action two-molecule propensity instead uses $x(x-1)$. The calculation requested here uses the printed formal generator and its large-copy-number approximation, where such boundary corrections are negligible.

The given Riccati ansatz is $v^{\prime }=-i{v}^2+iz/6$. Differentiating it and substituting for $v^{\prime }$ showsIt therefore selects $C=i/6$ in the preceding scaled Painleve equation. Write $v=-i{\Psi }^{\prime }/\Psi$. The quadratic terms cancel in the ansatz, leavingThus every nonzero solution of the indicated Airy equation gives the local mKdV solutionon intervals where $\Psi \ne 0$. For example $\Psi (z)=A\mathrm{Ai}(-z/6^{1/3})+B\mathrm{Bi}(-z/6^{1/3})$, with constants not both zero, has the required equation. This branch is generally complex and meromorphic; the printed ansatz itself includes $i$. A real $v$ on a real interval would require simultaneously $v^{\prime }=0$ and $v^2=z/6$, which is impossible on an interval. Hence this particular ansatz does not furnish a nontrivial real-valued mKdV profile.

The scaling of the equation suggests $q=t^{-1/3}v(z)$, $z=x{t}^{-1/3}$, for $t>0$. Substitution givesIntegrating in $z$ gives the scaled Painlevé II equationwhere $C$ is an arbitrary integration constant. If the standard normalization $w^{\prime \prime }=2w^3+sw+\alpha$ is desired, set $z=3^{1/3}s$ and $v(z)=i{3}^{-1/3}w(s)$. Then $\alpha =-3iC$ andThe direct scaled equation avoids introducing imaginary variables when a real $v$ is required.

For a localized traveling wave put $q(x,t)=Q(\xi )$ with $\xi =x-ct-x_0$, and require $Q,Q^{\prime },Q^{\prime \prime }\to 0$ at infinity. The modified KdV equation integrates once to $Q^{\prime \prime }-cQ+2Q^3=0$. Multiplying by $Q^{\prime }$ and integrating again givesA nonzero decaying real solution requires $c>0$. Separating on the rising or falling half of the wave and using the indicated inverse-hyperbolic-secant integral yieldsThe signs and translations are arbitrary. Differentiating the sech profile verifies $Q^{\prime \prime }=cQ-2Q^3$ directly. The relevant antiderivative is the inverse hyperbolic secant, since $d(\mathrm{arcsech}x)/dx=-1/[x\sqrt{1-x^2}]$ for $0<x<1$.

Set $r=p/q$. If $r\notin [1,2]$, then $|\sqrt{2}-r|\ge \sqrt{2}-1>1/4\ge 1/(4q^2)$. If $1\le r\le 2$, irrationality of the square root of two implies that the integer $2q^2-p^2$ is nonzero. Factoring a difference of squares therefore givesThus the requested bound holds for every integer $p$ and every nonzero integer $q$, without using the Liouville approximation theorem. For completeness, irrationality follows because $p^2=2q^2$ in lowest terms would make $p$ even and then $q$ even, contradicting coprimality.