For a flat matter-dominated universe, the Friedmann solution has $a\propto t^{2/3}$, $H=2/(3t)$ and $\bar{\rho }=1/(6\pi G{t}^2)$. Thus the density equation isTrying $t^s$ gives $s(s-1)+4s/3-2/3=0$, with roots $2/3$ and $-1$. ThereforeThe growing mode is proportional to the scale factor. When smooth dark energy or a cosmological constant dominates, the matter source density falls while expansion provides continued damping, suppressing growth. In the late constant-$\Lambda$ limit, $H\to H_{\Lambda }$ and $\bar{\rho }\to 0$, so $\ddot{\delta }+2H_{\Lambda }\dot{\delta }\simeq 0$ and $\delta$ tends to a constant plus a decaying $e^{-2H_{\Lambda }t}$ term. Matter perturbations therefore cease growing appreciably rather than continuing the matter-era law indefinitely.

A comoving material cell contains the same particles in its perturbed and background descriptions, so its mass is $\bar{\rho }{a}^3{d}^{3}q=\rho d^{3}r$. The Jacobian determinant is $a^3\det (I+{\nabla }_q\psi )$. Expanding the determinant givesSplit the potential into homogeneous and perturbed parts. There is a sign error in the printed hint: ${\nabla }^2{\Phi }_0=4\pi G\bar{\rho }$ requires $\nabla {\Phi }_0=+4\pi G\bar{\rho }r/3$, not the printed negative sign. Its acceleration $-\nabla {\Phi }_0$ agrees with $\ddot{a}/a=-4\pi G\bar{\rho }/3$.

For the perturbation, ${\nabla }_q^2\varphi =4\pi G\bar{\rho }{a}^2\delta =-4\pi G\bar{\rho }{a}^2{\nabla }_q\cdot \psi$. Under the usual homogeneous boundary convention the curl-free force therefore satisfieswhere ${\psi }_L$ is the longitudinal part of the displacement. Expanding the particle acceleration gives $\ddot{r}=\ddot{a}(q+\psi )+2\dot{a}\dot{\psi }+a\ddot{\psi }$. The background acceleration cancels the first term, leavingFor a longitudinal displacement this proves the displayed longitudinal displacement in linear dust perturbations equation with ${\psi }_L=\psi$. For unrestricted displacement, a transverse component has no density contrast and instead obeys ${\ddot{\psi }}_T+2H{\dot{\psi }}_T=0$; the full-vector equation printed in the question thus needs the longitudinal assumption. Taking the divergence removes this qualification and gives

Set $t=x^2$, so $dx=dt/(2\sqrt{t})$. ThenThe beta--gamma identity, the Gamma function recurrence and the gamma reflection formula give

Choose $F(z)=(z+1{)}^{1/3}{z}^{1/3}(z-1{)}^{1/3}$, each factor initially defined by polar argument in $(-\pi ,\pi )$. The products of the boundary values agree on $(-\infty ,-1)$, so their apparent cuts there cancel; the product extends analytically to $\mathbb{C}\setminus [-1,1]$ and has $F(z)\sim z$ at infinity. In polar coordinates about the three branch points this choice is $\prod r_j^{1/3}\exp [i\sum {\theta }_j/3]$.

For $0<x<1$, its upper and lower bank values are $e^{\pm i\pi /3}[x(1-x^2){]}^{1/3}$. For $-1<x<0$ they are $e^{\pm 2i\pi /3}[(-x)(1-x^2){]}^{1/3}$. The upper-minus-lower jump on each half is therefore $2i\sin (\pi /3)$ times the positive real integrand. Deforming a positively oriented large circle onto the cut, with the upper bank traversed right to left, givesSmall circles about the branch points contribute zero in the limit. At infinity,so the large-circle integral is $-2\pi i/3$. HenceBoth the branch choice and the orientation are essential to the sign.

For the plain pseudoprime assertion use the Fermat primality test. In the finite group of units modulo an integer, letIt is the kernel of a group homomorphism, because the unit group is an abelian group. The given witness makes $L$ a proper subgroup. By Lagrange's theorem for finite groups, $|L|\le \phi (N)/2$. Thus at least half of all units fail the test. Removing base one removes a passing base, so among the permitted $1<b<N$ there remain $\phi (N)-|L|$ failures out of $\phi (N)-1$ bases, still at least half.

The same argument works for the Euler–Jacobi test: its passing bases form the kernel of $b\mapsto b^{(N-1)/2}(b/N{)}^{-1}$. One must not assert that strong-test passing bases form a subgroup; the kernel argument above concerns the Fermat statement requested here.

If $N$ is prime, Fermat's little theorem gives $b^{2^{s}d}\equiv 1$. If $b^d\equiv 1$, the strong test already passes. Otherwise let $j\ge 1$ be the first index with $b^{2^{j}d}\equiv 1$. Its preceding value $u=b^{2^{j-1}d}$ satisfies $u^2=1$ and $u\ne 1$ in the field ${\mathbb{F}}_N$. Since $(u-1)(u+1)=0$ in a field, $u=-1$. The index $j-1$ lies between zero and $s-1$, so every coprime base passes the strong test when $N$ is prime.

When $a>g$, the signs reverse. The downward configuration has an exponentially growing disturbance, whereas near $\theta =\pi$ one obtains $\ddot{\epsilon }+(a-g)\epsilon /\ell =0$. Upward is stable; downward is unstable. The effective gravity in the descending lift points upward in this case.

Near $\theta =0$, the linearized equation is $\ddot{\theta }+(g-a)\theta /\ell =0$. When $a<g$ this is an harmonic oscillator, so the downward configuration is stable. Writing $\theta =\pi +\epsilon$ instead gives $\ddot{\epsilon }-(g-a)\epsilon /\ell =0$, with an exponentially growing solution. Downward is stable; upward is unstable.

For $rN>a$ and $p>0$, a nonzero infected equilibrium requires $S_{\ast }=a/r$. Using $a{I}_{\ast }=p{R}_{\ast }$ and population conservation givesEliminate $R=N-S-I$ before assessing stability, since perturbations must preserve the total population. The resulting Jacobian matrix at this endemic equilibrium isIts trace is negative and its determinant positive, so both eigenvalues have negative real part: the endemic equilibrium is locally asymptotically stable on the fixed-population plane. The full three-variable system has a neutral direction corresponding to changing the conserved population.

Writing $\kappa =rN-a>0$, as $p\to 0$ the trace is $-prN/a+O(p^2)$ and the determinant is $p\kappa$. Hence the discriminant $(\mathrm{tr}J{)}^2-4\det J=-4p\kappa +O(p^2)<0$. As $p\to \infty$, the trace is $-p-\kappa +O(p^{-1})$ and the determinant is again $p\kappa$, giving a positive discriminant $p^2+O(p)$. Both quantities are $O(p)$ in both limits, but the eigenvalues are complex for sufficiently small $p$ and real for sufficiently large $p$.

Since $S\le N$, the infected population satisfies $\dot{I}=(rS-a)I\le (rN-a)I$. Integration gives $I(t)\le I(0)e^{(rN-a)t}$. Hence $rN<a$ forces the infected population to decrease exponentially from the outset: there is no epidemic outbreak.

The Dahlquist equivalence theorem states that a linear multistep method is convergent for well-posed sufficiently smooth initial-value problems, with consistent starting values, if and only if it is consistent and zero-stable. Zero-stability means that every root of its first characteristic polynomial lies in the closed unit disk, with roots on the unit circle simple.

Here the characteristic polynomials areConsistency holds because $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2a$. The two remaining roots have product one. For $0<a<2$ they are distinct conjugate unit roots, and neither equals one. At $a=0$, the root one is triple; at $a=2$, the root minus one is double. Outside $[0,2]$, the remaining roots are real reciprocal roots and one has modulus greater than one. ThusThe order conditions give $c_2=0$ and $c_3=6-a$, which is nonzero throughout the convergent range. Therefore the order of convergence is exactly two, for starting approximations of matching accuracy.

Insert a smooth exact function into the multistep method and expand its residual about $t_n$. DefineThe residual is ${\sum }_{k\ge 0}{h}^k{c}_k{y}^{(k)}(t_n)/k!$. Therefore order at least $p$ is equivalent to $c_0=\cdots =c_p=0$. Applying the residual to the monomials $(t-t_n{)}^k$, $k\le p$, gives exactly those conditions. Linearity then makes them equivalent to exactness on every polynomial of degree at most $p$, as asserted. For exact rather than at-least order $p$, the next coefficient must not vanish.

If an $s$-step formula were exact for degree $2s+1$, use Hermite interpolation to choose $Q$ with $Q(t_{n+i})=0$ for $i<s$, $Q(t_{n+s})=1$, and $Q^{\prime }(t_{n+i})=0$ for every $i$. Its degree is at most $2s+1$. The purported identity would then read ${\rho }_s=0$, contradicting ${\rho }_s=1$. No $s$-step method has order $2s+1$ or higher.

A simple zero at $ip$ supplies a discrete scattering mode and hence an order-one, right-moving soliton component of speed $4p^2$. Its position depends on the associated norming constant and, for general scattering data, the appropriate scattering phase shifts. A simple zero alone does not specify its position, nor exclude additional zeros and therefore additional solitons.

To locate the continuous radiation, set $\xi =x/t$. The real-axis oscillatory phase is $t(2k\xi +8k^3)$, whose stationary points satisfy $2\xi +24k^2=0$. For $\xi <0$ there are two real stationary points $k=\pm \sqrt{-\xi /12}$, giving dispersive oscillatory radiation, ordinarily of order $t^{-1/2}$ for nondegenerate data. For $\xi >0$ there is no real stationary point, so radiation is smaller and a soliton dominates near its moving centre. At $\xi =0$ the stationary points merge and the natural transition scale is $x=O(t^{1/3})$, with Airy-type radiation of order $t^{-1/3}$. These statements assume the regular decaying scattering data implicit in the reconstruction problem; vanishing reflection can remove the radiation entirely. The discrete mode travels right, while the leading continuous radiation occupies left-going rays.

When $b=0$, put $D=e^{-2px+8p^{3}t-2x_0}>0$. Since $c=2ip{e}^{-2x_0}$, evaluation at $ip$ gives $N(ip)=1-DN(ip)$, henceDifferentiate the coefficient of $1/k$ with respect to $x$, using $D_x=-2pD$, and apply the reconstruction formula. The result isThe positive sign follows from the source's $-2i$ reconstruction convention. The soliton has amplitude $2p^2$, speed $4p^2$, and centre $x=4p^{2}t-x_0/p$.

Suppress $(x,t)$, put $R(k)=b(k)/a(k)$ and $C=c{e}^{-2px+8p^{3}t}N(ip)$. Removing the pole gives the additive jumpThe upper analytic function is $\mathcal{M}-1$ and the lower analytic function is $N(-k)-1$. The Cauchy integral formula and the Sokhotski–Plemelj formula solve this jump as the upper and lower boundary values of $(2\pi i{)}^{-1}{\int }_{\mathbb{R}}G(s)/(s-z)ds$. At the lower point $z=-k$, with $\Im k>0$, this gives $N(k)-1$.

The rational part can be evaluated by closing the contour upward: ${\int }_{\mathbb{R}}[(s-ip)(s+k){]}^{-1}ds=2\pi i/(k+ip)$. Thus the required linear integral equation isIt is supplemented by its value at $k=ip$, and follows under the usual decay or principal-value hypotheses of the stated Riemann-Hilbert problem.

A power option need not have bounded payoff, but all real powers of a positive lognormal stock have finite moments, so risk-neutral valuation still applies. The Gaussian moment-generating identity gives $\mathbb{E}[e^{n\sigma B_T}]=e^{n^2{\sigma }^{2}T/2}$. ThereforeFor $n=1$ this reduces to $S_0$, and for $n=0$ to the time-zero value $e^{-rT}$ of one unit paid at expiry.

With continuous dividend yield $D$, it is the discounted total gain from the stock, including dividends, that must be a martingale. The ex-dividend stock drift under the risk-neutral measure is therefore $r-D$. Replace $r-{\sigma }^2/2$ in the exponential representation by $r-D-{\sigma }^2/2$ and repeat the chain rule. The discount rate remains $r$, givingThe terminal payoff condition is unchanged.

Under the given risk-neutral measure, the option price is $V(t,S)=e^{-r(T-t)}\mathbb{E}[\varphi (S_T)\mid S_t=S]$. Thus $e^{-rt}V(t,S_t)$ is a martingale. This valuation also follows from delta hedging: cancel the Brownian exposure with $V_S$ units of stock, after which absence of arbitrage makes the locally riskless remainder earn $r$.

Set $f(t,b)=e^{-rt}V(t,S_0{e}^{\sigma b+(r-{\sigma }^2/2)t})$. The stated Brownian martingale identity and the chain rule giveThe drift vanishes because discounted option value is a martingale. Consequently the Black-Scholes equation isFor bounded measurable payoffs, the conditional-expectation solution is smooth for $t<T$; terminal values are interpreted in the usual limiting sense at continuity points, or in the bounded measurable pricing sense.

The proposal must connect the whole positive posterior support by moves with nonzero reverse probability. Use the full Metropolis–Hastings acceptance probability for asymmetric proposals; dropping the reverse proposal factor generally targets the wrong distribution. Compute likelihoods and acceptance ratios on a logarithmic scale to avoid overflow and underflow, and handle zero weights explicitly. A lazy step prevents periodicity when distributional convergence is needed.

Finite-chain convergence does not guarantee a useful run length: separated modes and narrow bottlenecks can make mixing extremely slow. Tune proposals to explore efficiently, inspect several starting points and chains, and assess autocorrelation and Monte Carlo uncertainty rather than treating correlated draws as independent observations. Discarding an initial segment can reduce initialization effects, but diagnostics do not prove that every posterior mode has been visited. Thinning is not required for correctness and often throws away useful information. The model and prior must themselves be specified correctly; an exact sampler from an incorrect posterior cannot repair the model.

Restrict to the positive-weight support. From a state $\theta$, propose ${\theta }^{\prime }$ according to $q(\theta ,{\theta }^{\prime })$ and accept with probabilityOtherwise stay at $\theta$. For distinct states, the accepted transition kernel satisfiesThus detailed balance holds, and summing over $\theta$ proves stationarity. Irreducibility of the accepted kernel makes this stationary distribution unique. Proposal irreducibility by itself is insufficient if proposed edges have zero reverse probability and are always rejected.

For convergence of the distribution from an arbitrary start, also require aperiodicity, which can always be supplied by making the chain lazy. In a finite irreducible aperiodic chain, some power $P^m$ has all entries positive. Its rows then share a fixed positive probability component, so repeated application contracts total variation distance geometrically. This proves that the chain's distribution approaches $\pi$.

The stronger fact needed for averages isOne can justify this directly by returns to a chosen state. Successive excursions are independent identically distributed cycles by the Strong Markov property. In a finite irreducible chain, return times have finite mean: from every state there is a path of uniformly bounded length and uniformly positive probability to the chosen state, giving a geometric tail bound. Apply the strong law of large numbers to cycle lengths and accumulated rewards. Their ratio converges; the mean visits per cycle, normalized by the mean cycle length, form a stationary distribution and hence equal $\pi$. The unfinished final cycle is negligible. This proves the displayed average law, even when the chain is periodic. Under these conditions, Metropolis–Hastings delivers the required posterior expectations.