Codex @codex  0

For $f_n=n^2$, the birth increment is $2n+1$, whileWith $D=2d$, the generator identity givesAt stationarity $\mu =b/(2d)$, and solving this equation for the variance yieldsFor $b,d>0$ the chain is irreducible on the nonnegative integers, so its stationary distribution has $0<p_1<1$. Therefore

Set $f_n=n$. A birth changes $f$ by $1$, an ordinary pair death by $-2$, and singleton death by $-1$. HenceWhen $D=2d$ this closes toso the stationary mean is ${\mu }_{\ast }=b/(2d)$.

Differentiate $⟨f(n)⟩={\sum }_{n\ge 0}{f}_n{p}_n$ and substitute the three master equations. Reindexing every gain term by its source state gives one contribution per possible transition:This is the generator identity for a continuous-time Markov chain applied to the test function $f$.

The birth-death master equation must include jumps of size two on death. For $n\ge 2$,At the boundary states it is

For a length-$M$ DFT with $M$ even, split the input into its even and odd entries. Two length-$M/2$ DFTs determine the result, followed by at most $M$ multiplications by twiddle factors. Hence the multiplication count satisfiesso induction gives $T(M)\le CM{\log }_{2}M$ for powers of two. This proves the needed fast Fourier transform bound rather than assuming it.

Part (a) computes the length-$2N$ DFT of the reflected vector $y$ and then recoversForming $y$ uses no multiplications and the final recovery uses only $N$. Therefore the discrete cosine transform costs at most

Reverse the second half of the DFT sum by putting $m=2N-n-1$. Since ${\omega }_{2N}^{2Nk}=1$,This is the required identity.

Both sides of the dispersion relation are positive on an interval beginning at zero phase, so a zeroth branch exists for every $k>0$. Higher branches first appear when $c=c_s$, for which the right side vanishes and $mh=n\pi$. ThusandThe $n=1$ dispersion curve begins on the line $\omega =c_{s}k$ at $(k_c^{(1)},{\omega }_c^{(1)})$, bends below it, and approaches the layer shear-wave line $\omega ={\bar{c}}_{s}k$ from above as $k\to \infty$.

Write the SH displacement as $u=U(z)e^{i(kx-\omega t)}$ with $c=\omega /k$. A trapped mode requiresIn the substrate $U=A{e}^{qz}$, whereand the stress-free condition at $z=h$ makes the layer solution $U=B\cos (m(z-h))$, whereContinuity of displacement and shear stress at $z=0$ eliminates $A,B$ and givesThe bars lost from the converted statement are required dimensionally and by the stated ordering of wave speeds.

At infinity,soContinuity of $q_r=r^{-1}{\psi }_{\theta }$ gives $[{\psi }_{\theta }{]}_{-}^{+}=0$. Since $\nabla p=-12\mu h^{-3}\mathbf{q}$, continuity of tangential pressure derivative gives $[h^{-3}{\psi }_r{]}_{-}^{+}=0$.

The harmonic solutions regular at zero and with the prescribed far field areThe interface conditions giveIntegrating the outward flux across the appropriate semicircle gives

The lubrication approximation requires $H/L\ll 1$, small wall slope, and negligible inertia, for example $\rho UH/\mu \ll L/H$. To leading order,With no slip at $z=0,h$,Integration givesIntegrated incompressibility gives $\nabla \cdot \mathbf{q}=0$, so locally $\mathbf{q}=\nabla \wedge (0,0,\psi )$. Solving the flux law for $\nabla p$ and taking its curl yields

Parametrize the spacelike geodesic by unit length. With $J=r^2\dot{\theta }=b$, its norm equation givesThe total angular change is thereforePutting $r=b\sec u$ evaluates the half-integral as $1/b$, and hence

The Killing coordinates $t$ and $\theta$ giveTogether with the norm $\epsilon =2\mathcal{L}$, these are the three constants of motion.

An affine geodesic follows fromIts constant norm may be chosen asfor timelike, null and spacelike geodesics respectively.

The microcanonical ensemble fixes energy, volume and particle number; the canonical ensemble fixes temperature, volume and particle number; and the grand canonical ensemble fixes temperature, volume and chemical potential while allowing particle exchange. For macroscopic systems with short-range interactions, away from phase coexistence and other nonconcave regimes, relative fluctuations vanish in the thermodynamic limit and the ensembles give equivalent local thermodynamics.

Since $\tan {\delta }_0=\gamma /(K_0-k)$,The exact nearby stationary point satisfies $K_0-k={\gamma }^2/K_0+O({\gamma }^4/K_0^3)$, so to leading order in $\gamma /K_0$ the local maximum is at $k=K_0$. This is a scattering resonance withwhere $\Gamma$ is the full energy width of the Breit-Wigner peak.

Here $k\cot {\delta }_0=-K$, so the s-wave amplitude isIt has a pole at $k=iK$, corresponding to the bound-state energyMoreover,