When the tube radius is $a$ and the sphere radius is $a(1-ϵ)$, the minimum gap is $aϵ$ and its leading neck profile is $h=aϵ+z^2/(2a)$. Substitution $z=a\sqrt{2ϵ}t$ in $I_n^{\ast }=a^{n-1}\int h^{-n}dz$ gives the displayed coefficients. Constants change if the gap parameter is instead twice the radius deficit; the divergent powers do not.

For $h=H_0(1-x^2/a^2{)}^{1/6}$, elliptic coordinates make $h^3/\sin \eta$ constant on the upper crack face. Pressure continuity and harmonicity reduce the boundary condition to $F^{\prime }(0)=\alpha F(0)$ for the cosine mode, while uniform far flow gives $F\sim -e^{\xi }/2$. Solving $F^{\prime \prime }=F$ yields the displayed pressure. The midpoint crack flux is $2aU\alpha /(1+\alpha )$; small $\alpha$ leaves the background pressure gradient nearly unchanged, while large $\alpha$ makes the crack nearly equipotential.

For a bounded measurable function with positive essential supremum on a finite measure space, put $M_n=\int |f{|}^n$. The Cauchy-Schwarz inequality makes these moments log-convex, so successive ratios are nondecreasing. Their product satisfies $M_n/M_0\le (M_{n+1}/M_n{)}^n$, while each ratio is at most the essential supremum. Positive-measure level sets arbitrarily close to that supremum make $\mu (E{)}^{-1/n}\Vert f{\Vert }_n$ converge to it. Squeezing then proves the ratio limit.

Suppose $1\le p<q<r<\infty$, $\Vert f{\Vert }_p\le C_p$, $\Vert f{\Vert }_q\ge C_q>0$ and $\Vert f{\Vert }_r\le C_r$. Set $\epsilon =(C_q^q/(2C_p^p){)}^{1/(q-p)}$. The part of $\int |f{|}^q$ where $|f|\le \epsilon$ is at most ${\epsilon }^{q-p}{C}_p^p=C_q^q/2$. On the complementary level set, the Holder inequality bounds the integral by $C_r^q\mu \{|f|>\epsilon {\}}^{1-q/r}$, proving the bound. For $r=\infty$ the bound becomes $C_q^q/(2C_r^q)$. The lower-exponent bound prevents spreading into low amplitudes, while the higher-exponent bound prevents concentration into arbitrarily narrow spikes.

For $n\ge 2$ risk-neutral bidders with independent uniform distributions on $[0,1]$, the lowest-price auction has a symmetric Bayesian Nash equilibrium $b(v)=(n-1)v$. A deviation bidding $(n-1)z$, with $0\le z\le 1$, wins with probability $z^{n-1}$ and has expected payment $(n-1)z^n/n$. Its expected profit is $v{z}^{n-1}-(n-1)z^n/n$, whose derivative is $(n-1)z^{n-2}(v-z)$, proving optimality at $z=v$. Bids above $n-1$ give the same outcome as $z=1$. The seller receives $(n-1)$ times the minimum of $n$ independent uniform valuations, with expected revenue $(n-1)/(n+1)$.

// Target: mathematical-optimization.bigb

The concave biconjugate is an upper envelope of $F$, and equals $F$ for a proper closed concave function under the usual duality hypotheses. It equals $-(-F{)}^{\ast \ast }$ in terms of the ordinary biconjugate. If $e=D{F}_{\ast }(\tau )$, differentiability and concavity of $F_{\ast }$ imply $F_{\ast }(\tau )+F_{\ast \ast }(e)=\tau :e$. For a finite differentiable dual the equality follows because $\tau$ minimizes $s:e-F_{\ast }(s)$.

A lower bound in a partially ordered set which is greater than or equal to every other lower bound. For a pair it is written $a\wedge b$; an empty meet, when it exists, is the greatest element. In a complete lattice, every subset has a meet. In the open-set frame, this is the interior of the intersection, with finite meets given by the intersections themselves.

For a slow elastic layer above a faster half-space, put $q=\sqrt{c^2/{\bar{c}}_s^2-1}$ and $q_{\max }=\sqrt{c_s^2/{\bar{c}}_s^2-1}$. The positive branches of $\tan (khq)$ intersect the positive strictly decreasing traction ratio once whenever their left endpoint $m\pi$ is strictly below $kh{q}_{\max }$. The fundamental exists for every positive $kh$; the second requires the displayed strict inequality. At equality $c=c_s$ and the half-space decay exponent vanishes, so that endpoint is a cutoff, not a trapped mode satisfying decay at minus infinity.

For a slower surface layer supporting propagating SH-waves with vertical wavenumber $q>0$ above an evanescent half-space with decay exponent $p>0$, continuity of displacement field and traction gives the displayed reflection coefficient for a downward incident displacement field wave. Its modulus is one and its wave phase is $-2\arctan (\mu p/({\mu }_{0}q))$. A traction-free top reflects displacement field with coefficient $+1$. The round-trip condition $R_h{e}^{2iqh}=1$ then gives $\tan (qh)=\mu p/({\mu }_{0}q)$, equivalent to direct Love wave mode matching.

For a slow uniform layer over a faster half-space, the nth Love wave branch reaches the half-space shear wave speed at the displayed threshold. Its half-space decay exponent then vanishes, so the threshold is a limiting nonlocalized state, not a trapped finite-energy mode. The fundamental has zero threshold and exists at every positive frequency; overtones require frequency strictly above their positive threshold. At high frequency each fixed branch approaches the layer shear wave speed from above.

A free top and rigid base impose transverse wavenumbers $(n+1/2)\pi /h$ on guided SH modes between rigid and free planes. Their phase velocity and group velocity are $c={\beta }^{\prime }/\sqrt{1-{\omega }_n^2/{\omega }^2}$ and $U={\beta }^{\prime }\sqrt{1-{\omega }_n^2/{\omega }^2}$ for $\omega >{\omega }_n$. A real Love wave replaces the rigid base by an evanescent elastic load; the approximation becomes reliable sufficiently above the corresponding strong-dispersion transition, not at the actual mode cutoff.

For an aligned anisotropic surface layer, $q^2=({\omega }^2-{\beta }_x^{\prime 2}{k}^2)/{\beta }_z^{\prime 2}$, while the isotropic substrate has ${\kappa }^2=k^2-{\omega }^2/{\beta }^2$. Zero top traction and welded displacement/traction continuity give the displayed dispersion relation. The horizontal speed sets the limiting phase velocity; the vertical speed sets rigid-base cutoff frequencies. The effective contrast for the high-contrast transition is ${\rho }^{\prime }{\beta }_x^{\prime }{\beta }_z^{\prime }/(\rho {\beta }^2)$.

Three nodes have integer link capacities $C_{12},C_{13},C_{23}$. Calls between a node pair use either its direct link or the two-link alternative. With unrestricted instantaneous rearrangement of existing calls, counts $n_{12},n_{13},n_{23}$ are feasible exactly when $n_{12}+n_{13}\le C_{12}+C_{13}$, $n_{12}+n_{23}\le C_{12}+C_{23}$ and $n_{13}+n_{23}\le C_{13}+C_{23}$. Necessity follows by counting calls crossing each node cut. For sufficiency, the inequalities allow at most one count to exceed its direct-link capacity. Divert precisely that excess to the other two links; the same inequalities ensure capacity there. The counts thus have the same feasible set as a fixed routing loss network with one virtual resource at each node, capacity equal to the sum of its two incident physical-link capacities, and each call using its two endpoint resources. With independent Poisson processes for arrivals and independent routing-independent exponential distributions for holding times, this gives the usual product-form stationary distribution of a loss network and exact Poisson arrivals see time averages blocking probabilities.

Here ${\mathcal{R}}_{\mathrm{lc}}=J_{\mathrm{lc}}^2/J_c^2$ is the orbital loss cone fraction, $P$ an orbital period, and $t_{\mathrm{rel}}$ a stellar relaxation time. When angular diffusion per orbit is small relative to the cone, a steady two-dimensional diffusion solution is logarithmic and gives the empty-cone flux. Rapid refilling maintains an approximately isotropic full cone and gives the second rate. Numerical total rates need a stellar density and orbital-energy distribution, not merely a star count.

For a homologous black hole-dominated cluster, write $E=-bG{M}_{H}Nm/(2R)$, and let a captured star's specific orbital energy be ${ϵ}_{\mathrm{cap}}=\xi E/(Nm)$. Removing negative orbital energy raises the remaining cluster energy, while growth ${\dot{M}}_H=m\Gamma$ deepens its potential. The energy budget gives the displayed radius response. Preferentially removing strongly bound stars can drive expansion; representative-energy removal instead gives the adiabatic contraction due to black-hole growth. The sign is not fixed by the loss cone angle alone.

The magnetic part of the Lorentz force is perpendicular to the particle velocity. Its power is therefore zero, so with no other forces the kinetic energy remains constant. This remains true relativistically because $dE/dt=\mathbf{F}\cdot \mathbf{v}$; a magnetic force can change momentum direction without changing total energy.

For a particle of positive mass $m$ and signed electric charge $q$ moving under the magnetic-only Lorentz force in $\mathbf{B}=B_z(t){\mathbf{e}}_z$, write $u=\dot{x}+i\dot{y}$. The planar equation isThus the planar speed is constant: a changing magnetic field changes the rate of rotation without doing work in this prescribed force model. The axial velocity is constant as well.

For $B_z(t)=\beta t$, initial planar velocity $(V,0)$ with $V>0$, and zero planar initial position,These are Fresnel integrals. If $q\beta \ne 0$, their limits areThe negative second coordinate therefore assumes $q\beta >0$; reversing the electric charge or the ramp reverses the transverse deflection. When $q\beta =0$ the motion is ballistic and has no finite position limit. The finite limit for a nonzero ramp is compatible with a nonzero speed: the turns shrink as their angular rate increases without bound.

A physical changing magnetic field also obeys Faraday's law of induction. An induced electric field can do work and modify this motion, so these expressions describe the explicitly magnetic-only model rather than a complete experimental ramp.

// Target: number-theory.bigb

For an axial magnetic field $\mathbf{B}=(B_0/r){\mathbf{e}}_z$, let $L=m{r}^2\dot{\theta }$. The Lorentz force gives $\dot{L}=-q{B}_0\dot{r}$, so $\ell =(L+q{B}_{0}r)/m$ is constant. Since magnetic forces do no work and $\dot{z}$ is constant, the transverse speed $v_{\perp }$ is constant and ${\dot{r}}^2=v_{\perp }^2-(\ell /r-c{)}^2$. If $c\ell >0$ and $0<v_{\perp }<|c|$, two finite radial turning points confine the motion. If $v_{\perp }>|c|$, an outward branch escapes to infinite radius. At equality a marginal escaping branch exists when $c\ell >0$, but $\ell =0$ gives a circular orbit instead. The transverse escape velocity threshold is $|q{B}_0|/m$; motion parallel to the magnetic field cannot provide transverse escape energy.

For particles with the same charge-to-mass ratio, rotationally invariant mutual central forces, and a weak uniform magnetic field, a frame with the displayed angular velocity cancels the velocity-dependent magnetic Lorentz force against the Coriolis force. The remaining rotational and magnetic terms are quadratic in $\mathbf{B}$. Thus, to first order in the field, the rotating-frame equations coincide with the zero-field equations. This is the classical dynamical basis of Larmor precession; equality is an approximation rather than an exact removal of the magnetic field.

The gyroradius is the radius of the transverse circular motion of a charged particle in a uniform magnetic field. In nonrelativistic classical mechanics, with zero electric field and no other forces, it is $m{v}_{\perp }/(|q|B)$, where $v_{\perp }$ is the speed perpendicular to the magnetic field. The parallel component of velocity changes the advance of the helix, rather than its transverse radius.