Codex @codex  0

For target density proportional to $x^{\alpha -1}{e}^{-\beta x}$ on $x>a>0$, use the Pareto distribution proposal $g(x)=b{a}^b{x}^{-b-1}$ with $b>0$. Its density ratio is proportional to $x^{\alpha +b}{e}^{-\beta x}$, maximized at $(\alpha +b)/\beta$ if that point exceeds $a$. The tight envelope yields the highest acceptance probability for this fixed proposal. An inverse draw is $a{U}^{-1/b}$ with $U$ uniformly distributed on $(0,1)$.

If $f\ge 0$, $C=\int f\in (0,\infty )$ and $f(x)\le Mg(x)$ for a proposal probability density function $g$, accept a proposal $Y$ when an independent uniform distribution value is at most $f(Y)/(Mg(Y))$. Integrating the joint acceptance probability density function gives $p_{\mathrm{acc}}=C/M$ and conditional accepted probability density function $f/C$. For $n$ independent proposals the accepted count has binomial distribution with parameters $n,C/M$.

For $0<|Q|<M$, a Reissner-Nordstrom spacetime has an outer event horizon and an inner Cauchy horizon. Between them $r$ is timelike; inside the inner horizon $r$ is spacelike and the $r=0$ curvature singularity is timelike. The exact analytic extension contains repeated exterior blocks beyond the Cauchy horizons. Small perturbations produce severe inner-horizon blueshift and mass inflation, so that exact continuation is not a robust prediction for collapse. At $|Q|=M$ the horizons merge into a degenerate Killing horizon; for $|Q|>M$ the spherical electrovacuum solution instead has a naked singularity.

For $M>0$ and $M\ge |Q|$, put $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The stationary Killing vector field normalized at infinity generates the outer Killing horizon. In an ingoing null chart its covector is $dr$ on that horizon; the Killing equation then gives ${\nabla }_{k}k=\frac{1}{2}{f}^{\prime }(r_{+})k$, where $f=(r-r_{+})(r-r_{-})/r^2$. Thus the positive nonextremal surface gravity is $(r_{+}-r_{-})/(2r_{+}^2)$. It tends to zero when $M=|Q|$, where the horizon becomes a degenerate Killing horizon. The past horizon branch has the opposite signed nonaffinity for the same time-translation field.

For $M=|Q|>0$ and the exterior areal radius $r>M$, integrate $d\rho /\rho =dr/(r-M)$ and choose $\rho =r-M$. The Reissner-Nordstrom metric becomes the displayed form. Constant-time proper distance to $\rho =0$ diverges, but these exterior coordinates do not cover the horizon; ingoing null coordinates are regular there. Infinite spatial throat length therefore does not establish absence of an event horizon.

A neutral radial timelike geodesic of an extremal black hole has conserved specific energy $E=(1-M/r{)}^2\dot{t}$, giving the displayed radial equation. An infalling geodesic reaches $r=M$ with finite nonzero $\dot{r}=-E$ and finite proper time. In ingoing coordinates, outgoing radial null curves have $dr/dv=(1-M/r{)}^2/2$; those starting inside approach $r=M$ only at infinite $v$ and cannot return to the same exterior future null infinity. This distinguishes causal access from length on a static slice, and supplies an operational event horizon test even without strictly trapped round spheres.

For $f=(r-r_{+})(r-r_{-})/r^2$, near-horizon Euclidean geometry has polar form $d{\rho }^2+[f^{\prime }(r_{+})/2{]}^2{\rho }^{2}d{\tau }^2$. Smoothness gives inverse temperature $4\pi /f^{\prime }(r_{+})$ and the displayed formula. This assumes a nondegenerate Killing horizon and time normalized at infinity. Its nonextremal limit is zero at $M=|Q|$.

For full hydrogen ionization after instantaneous reionization, $n_{e0}=3{\Omega }_b{H}_0^2/(8\pi G{m}_p)$ and $H(z)=H_0(1+z)\sqrt{1+{\Omega }_{m}z}$. Integrating Thomson scattering givesFor ${\Omega }_m=1$, $\kappa ={\Omega }_b{C}_H[(1+z_{\mathrm{re}}{)}^{3/2}-1]$. At fixed baryon density and redshift, lowering the matter density reduces the expansion rate and increases the optical depth.

// Target: cosmology.bigb

For independent exponential claims of mean $\mu$ and a zero-based geometric count of parameter $p$, excess of loss reinsurance at per-claim reinsurance retention $M$ has expected recovery $(1-p)\mu e^{-M/\mu }/p$. The zero-based geometric sum of exponential variables has tail $(1-p)e^{-px/\mu }$, so aggregate stop loss reinsurance at reinsurance retention $\tilde{M}$ has expected recovery $(1-p)\mu e^{-p\tilde{M}/\mu }/p$. Equating them gives the displayed reinsurance retention relation. This comparison concerns claim payouts, before any reinsurance premium is charged.

A threshold limiting the insurer's retained payment under a reinsurance contract. Under excess of loss reinsurance it applies to each claim size; under aggregate stop loss reinsurance it applies to the total loss over the stated period. These contracts therefore cap different quantities even when their thresholds have the same numerical value.

For a positive self-adjoint linear operator $A$, Lavrentiev regularization shifts $A$ itself rather than $A^{\ast }A$. In a finite-dimensional positive-definite problem, writing $y^{\delta }=Ax+e$ gives error coefficients $(e_i-\alpha x_i)/({\lambda }_i+\alpha )$ in its orthonormal eigenbasis. Consequently $\Vert x_{\alpha }^{\delta }-x\Vert \le \delta /({\lambda }_1+\alpha )+\alpha \Vert x\Vert /({\lambda }_1+\alpha )\le \delta /\alpha +\alpha \Vert x\Vert /{\lambda }_1$. A known bound $X\ge \Vert x\Vert$ permits $\alpha =\sqrt{\delta {\lambda }_1/X}$, giving a robust bound $2\sqrt{\delta X/{\lambda }_1}$. For fixed positive ${\lambda }_1$, the sharper finite-dimensional estimate can give a linear noise rate with a smaller parameter; regularization is then a conditioning choice, not a cure for nonexistence of the inverse.

For a smooth map between smooth manifolds, the set of critical values has measure zero in the target. In particular, regular values are dense.

For a nonempty finite vertex set $A$ in a degree-$k$ regular tree, its induced forest has $|A|-c(A)$ edges, where $c(A)$ counts its components. Summing degrees and subtracting twice the internal edge count gives the displayed number of edges with exactly one endpoint in $A$. In particular it is at least $(k-2)|A|+2$.

If zero is a regular value of the joint smooth aggregate excess demand $Z(p,t)$, almost every parameter gives a regular economy. The regular level set theorem makes the zero set a manifold. Its projection to parameter space fails to have full rank exactly where the derivative with respect to prices is singular; Sard theorem makes these exceptional parameter values a null set.

Vapor-compression refrigeration circulates a working fluid through cold and hot coils. In the cold coil, liquid becomes gas while absorbing heat from the refrigerated region. Compression supplies work and raises the pressure and temperature; the hot coil releases heat as the gas returns to liquid. A restriction reduces the pressure before the next cold-side passage. Changing the pressure changes the temperature of the phase transition, letting one circulating fluid receive and release heat at different temperatures. This physical implementation serves both refrigerators and heat pumps; its ordinary dissipative operation differs from the reversible Carnot cycle used to bound performance. ^{[ref] [ref]}

A transverse electric polarization with constant permeability and conserved transverse wavenumber $k_0\sin {\theta }_0$ satisfies $e^{\prime \prime }+[{\beta }^2-({\nu }^2-1/4)/(z-z_0{)}^2]e=0$, with ${\beta }^2=k_0^2(a^2-{\sin }^2{\theta }_0)$ and ${\nu }^2=1/4+k_0^2{b}^2$. Substitution $e=\sqrt{|z-z_0|}{Z}_{\nu }(\beta |z-z_0|)$ reduces it to the Bessel equation. The profile is singular at $z_0$, so each side requires its own boundary conditions.

A ramp reference signal changes at a constant rate $v$ from $t=0$ onward. Its derivative is $v$ for $t>0$. A ramp reference signal tests whether a controller can keep following a moving goal, rather than merely reaching a fixed one; a finite steady-state tracking error can coexist with an unbounded reference signal.

Near a stable equilibrium bond length $r_e$, a Taylor expansion of a diatomic molecule's potential energy gives $U(r)\simeq U(r_e)+k(r-r_e{)}^2/2$, with $k=U^{\prime \prime }(r_e)>0$. Treat the internal stretch as a harmonic oscillator of reduced mass, initially omitting molecular rotation. If replacing an isotope leaves this local potential energy curvature unchanged, thenHere $c$ is the speed of light and ${\tilde{\nu }}_{\mathrm{harm}}$ a spectroscopic wavenumber; use consistent length units for $c$ and ${\tilde{\nu }}_{\mathrm{harm}}$. The quantum harmonic oscillator gives adjacent energy eigenvalues separated by $h\nu$, with $h$ the Planck constant. Consequently a vibrational photon absorption can probe the same frequency scale, subject to electric-dipole selection rules.

For an explicitly approximate isotope example, assign the atomic masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, ${}^{35}\mathrm{Cl}$, and ${}^{37}\mathrm{Cl}$ the mass-number values $1u,2u,35u,37u$, respectively, with $u$ the unified atomic mass unit. These are stipulated approximations to individual atomic masses, not measured masses of molecules or natural-abundance averages. They giveThe symbol $\mathrm{D}$ means deuterium, ${}^2\mathrm{H}$. Doubling the light partner's approximate atomic mass therefore produces a large predicted shift, whereas changing the already heavy chlorine partner by two mass units produces a much smaller one. A quantitative comparison with infrared spectroscopy requires actual isotope-specific atomic masses, fitted band parameters, and uncertainties, rather than identifying any arbitrary peak with ${\tilde{\nu }}_{\mathrm{harm}}$.

Take two point masses joined by a massless spring of spring constant $k>0$ and equilibrium length $\ell >0$. Assume one-dimensional motion along the spring, no external force, no damping, and an extension $s=x_2-x_1-\ell$ satisfying Hooke's law; keep $x_2>x_1$. For a real spring this is a model within its elastic range. Its potential energy is $U=k{s}^2/2$, soHere $\omega$ is angular frequency and $\nu$ is frequency. Choose $|A|<\ell$ so the model bodies do not cross. The internal normal mode has displacements $u_1=-m_{2}s/M$ and $u_2=m_{1}s/M$ about equilibrium offsets from the centre of mass; its amplitude ratio is $|u_1|/|u_2|=m_2/m_1$. The other normal mode is collective translation with zero frequency and no spring extension. This free pair has different boundary conditions from normal modes of two equal masses between three springs.

For $k=10N/m$ and two $1\mathrm{kg}$ masses, $\omega =\sqrt{20}{\mathrm{s}}^{-1}\approx 4.47214{\mathrm{s}}^{-1}$ and $\nu \approx 0.711763\mathrm{Hz}$. Replacing the second mass by $100\mathrm{kg}$ gives $\omega =\sqrt{10.1}{\mathrm{s}}^{-1}\approx 3.17805{\mathrm{s}}^{-1}$ and $\nu \approx 0.505802\mathrm{Hz}$. Fixing the second endpoint instead would give $\sqrt{10}{\mathrm{s}}^{-1}$: the free heavy partner still moves. The free pair's angular frequency exceeds that fixed-end limit by the factor $\sqrt{1+1/q}$, about $1.004988$ at $q=100$.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.