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The word "projection" does not mean ordinary perpendicular projection in the plotted Euclidean space. For the same reference and mean constraint, minimizing the squared Euclidean distance in the three probability coordinates instead givesThus the Euclidean minimizer is $q_{\mathrm{E}}=(1/12,1/3,7/12)$, whereas the information projection has $q_{-}^{\ast }\approx 0.116204$. The squared Euclidean objective at its own minimum is $1/8$; its relative entropy is approximately $0.210918$ nats, larger than the optimum $0.197378$ nats. Both distributions satisfy the same constraint.

The reference and objective encode what "least change" means. Relative entropy is asymmetric and is not a metric; its minimization answers a different question from linear least squares. The useful geometric analogy comes from a Pythagorean inequality for information projection, rather than from an assertion that the colored contours are circles.

For any real constant $c$, one-dimensional Brownian motion repeatedly returns to zero. Consider successive segments from zero until exiting $(-1,1)$, separated by returns to zero. Their capped durations are independent identically distributed positive bounded random variables by the Strong Markov property, so their sum diverges by the strong law of large numbers. On each such segment $e^{c{B}_s}\ge e^{-|c|}$. Thus the exponential clock has infinite lifetime, despite arbitrarily deep negative excursions.

Choose a maximal Brauer pair $(D,b_D)$ for the block. Its inertial quotient is $N_G(D,b_D)/(D{C}_G(D))$, and its order is the inertial index. For cyclic $D$ of order $q$, this is a prime-to-$p$ divisor of $p-1$. The number of edges of the block's Brauer tree is this index, and the exceptional multiplicity is $(q-1)/e(B)$.

For $N◃G$, $\theta \in \mathrm{Irr}(N)$ and $T=I_G(\theta )$, induction gives the displayed bijection. The inverse takes the $\theta$-isotypic component of an irreducible $G$-module, viewed as a $T$-module. Induction produces distinct $N$-isotypic components indexed by $G/T$. Any nonzero $G$-submodule meets one of them and hence, by translation and irreducibility over $T$, contains them all. The same argument shows that the inverse component is irreducible and determines the original inducing module uniquely.

For two independent uniform valuations on $[0,1]$, optimal posted price revenue is $2/(3\sqrt{3})$, attained at $p=1/\sqrt{3}$. An English auction without reserve earns the expected minimum, $1/3$. With reserve $r\in [0,1]$, revenue is $r\mathrm{Pr}(\text{exactly one valuation exceeds }r)$ plus the expected smaller valuation when both exceed $r$, namely $1/3+r^2-4r^3/3$. It is maximized at $r=1/2$, giving $5/12$. Different reserve-dependent allocation probabilities explain why this differs from the no-reserve revenue without violating revenue equivalence.

The independent private values model is symmetric when bidders have identical roles and draw their valuations independently from the same distribution. For revenue equivalence in the usual single-object setting, bidders additionally have risk neutrality and quasilinear utility. Under a strictly increasing symmetric equilibrium bidding rule, a type $v$ wins with probability $F(v{)}^{n-1}$.

// Target: mathematical-optimization.bigb

For a horizontal field $B(z)e_x$ in constant downward gravity, equilibrium obeys $(p+B^2/(2{\mu }_0){)}_z=-\rho g$. The displacement energy reduces to the displayed quadratic form. Perturbations independent of the field direction do not bend magnetic field lines and therefore have no magnetic tension penalty: an increasing mass density with height permits instability. A nonincreasing mass density gives nonnegative energy for all admitted incompressible perturbations. Along-field variation adds magnetic tension proportional to wavenumber squared, stabilizing sufficiently short waves when buoyancy is unfavorable.

A projective bundle with fibre ${\mathbb{P}}^{15}$ over the four-dimensional Grassmannian $\mathrm{Gr}(2,4)$. The condition of containing a fixed line imposes four independent linear conditions on the twenty cubic coefficients. The projection to ${\mathbb{P}}^{19}$ is proper. The line $\{x_2=x_3=0\}$ is an isolated point over $x_0^2{x}_2+x_1^2{x}_3$; isolated fibre point forces dominance in equal dimensions then proves that every cubic surface contains a line.

For an initial mass function $dN\propto m^{-\alpha }dm$ on $0.1<m/M_{\odot }<100$, complete loss of material initially above $10M_{\odot }$ removes fractionThe impulsive disruption by stellar mass loss threshold is $F=1/2$, with unique finite root $\alpha \simeq 1.79101236$. Decreasing $\alpha$ shifts mass to larger stars and increases $F$, so the idealized system is globally unbound for $\alpha <{\alpha }_{\mathrm{crit}}$. The derivation requires rapid, well-mixed loss and absence of a confining external potential.

In the local incompressible weak-field limit, take an axisymmetric wavevector $(k_r,0,k_z)$ and ${\omega }_A=\mathbf{k}\cdot \mathbf{B}/\sqrt{{\mu }_0\rho }$. Eliminating the pressure projects the radial force by $k_z^2/k^2$. This yields the displayed dispersion relation, where $\kappa$ is the radial epicyclic frequency. For a hydrodynamically stable rotation profile with outward-decreasing angular velocity, instability requires $0<{\omega }_A^2<-(k_z^2/k^2)d{\Omega }^2/d\ln r$. A purely toroidal field has no axisymmetric local field-line-bending frequency and therefore no ordinary weak-field MRI in this approximation.

For local cylindrical shear $q=-d\ln \Omega /d\ln r$ with $0<q<2$, maximizing the positive squared-growth root of the ideal magnetorotational dispersion relation gives the displayed formulas. For a Keplerian disk these are $s_{\max }=3\Omega /4$ and ${\omega }_A^2=15{\Omega }^2/16$. Field strength selects the fastest wavelength, while the maximum ideal rate is set by shear. Finite domains and magnetic diffusion can exclude that wavelength.

For an ideal $I\subseteq R$, the Rees ring is the graded subringIt packages all powers of $I$ into one algebra and turns stable filtrations into finitely generated graded modules.

Restrict a cubic equation to the affine line through two distinct singular points. At each point the equation and its first directional derivative vanish, so the restricted polynomial has two distinct double roots. A polynomial of degree at most three with this property is identically zero. Hence the whole line lies in the hypersurface. The multiplicity argument works in every characteristic.

An X-piece is a curvature-$-1$ four-holed sphere with geodesic boundary, formed by gluing two Y-pieces along equal-length cuffs. For a specified separating cuff and labelled outer boundaries, four outer lengths, the internal cuff length $s>0$, and a twist $\tau$ determine the metric. The gluing is orientation-reversing on boundary circles, with relative arclength displacement $\tau$ measured from chosen seam endpoints. Without markings, the gluing displacement is taken modulo $s$; with a marking, $\tau \in \mathbb{R}$ records Dehn twists as well. Changing the chosen separating curve can give different parameter descriptions of the same unmarked surface.

For a fixed positive integer $q$, $\cosh (1/q)={\sum }_{k\ge 0}{q}^{-2k}/(2k)!$ is irrational. If it were rational with denominator $b$, choose $N$ large enough that $b\mid (2N)!$. After multiplication by $q^{2N}(2N)!$, the finite partial sum is an integer and the remaining positive tail is bounded by a geometric series whose ratio is at most $1/[q^2(2N+1)(2N+2)]$. For sufficiently large $N$, this tail is strictly between zero and one, contradicting its integer value. Only a finite integer partial sum is used; convergence of an infinite series of rationals does not itself prove irrationality.

A catenary is the static shape of an ideal flexible, inextensible chain hanging under its own gravitational weight in a uniform gravitational field, with uniform linear mass density along its arc length. With the lowest point above $x=0$, one convenient vertical origin givesHere $a$ has units of length. The supports and the chain's arc length determine the scale and position; horizontal and vertical shifts give the general catenary. ^[ref]

For $a=1$, the three points at $x=-1,0,1$ have heights $\cosh 1\approx 1.543081$, $1$, and $\cosh 1\approx 1.543081$. If heights should instead be measured from the lowest point, subtract $a$:The same three heights are then approximately $0.543081$, $0$, and $0.543081$. The shift changes the vertical origin and preserves the catenary shape.

The scale also measures tension: if $H>0$ is its constant horizontal component, $\mu$ is the linear mass density, and $g$ is the magnitude of gravitational acceleration, then $a=H/(\mu g)$. A short force balance explains this. For the portion from the lowest point to $x\ge 0$, with arc length $s(x)$, vertical tension balances its gravitational weight and horizontal tension givesApplying differentiation yields $a{y}^{\prime \prime }=\sqrt{1+y^{\prime 2}}$. The proposed hyperbolic cosine solves this differential equation, because $y^{\prime }=\sinh (x/a)$ and $a{y}^{\prime \prime }=\cosh (x/a)=\sqrt{1+{\sinh }^2(x/a)}$. This is the mechanism behind the shape, under the stated ideal chain assumptions.

The same identity simplifies the arc length between symmetric supports at $x=\pm b$:For $a=b=1$, $L\approx 2.350402$, exceeding the horizontal span of two. A larger $a$ at fixed $b$ gives a shallower sag: the Taylor series gives $a(\cosh (x/a)-1)=x^2/(2a)+x^4/(24a^3)+\cdots$. Thus a parabola is a local approximation, while the exact ideal chain shape is a catenary. The useful connection is that the same hyperbolic Pythagorean identity controls its coordinates, force balance, and arc length.

For $g^{\prime }(0)\ne 0$, apply Jensen's formula with the smooth curvature term to the hyperbolic derivative density. Its distributional Laplacian is $H_g^2+2\pi \sum m_a{\delta }_a$, giving the displayed identity with ramification points defined by $g^{\prime }(a)=0$. If $g^{\prime }(z)=c{z}^m+\cdots$ at zero, use $m\log R$ as the origin counting term and replace the base constant by $\log {\lim }_{z\to 0}{H}_g(z)/|z{|}^m$. Without this regularization the basepoint-critical formula is undefined.

In a grey atmosphere with $T^4=T_0^4(1+3\tau /2)$ and $\kappa ={\kappa }_0{P}^{\alpha -1}{T}^{4-4\beta }$, hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. With $\alpha ,\beta >0$ and $P(0)=0$, integration gives the displayed pressure-temperature relation. At the matching surface $T=T_e$, the Eddington surface boundary condition gives $T_e^4=2T_0^4$, hence $P\kappa /g=4\alpha (1-2^{-\beta })/(3\beta )$.

Take an ideal gas grey atmosphere with $T^4=T_e^4(3\tau +2)/4$, constant surface gravity of a star $g$, vanishing external pressure and $\kappa ={\kappa }_0{P}^{1/2}{T}^8$. Integrating hydrostatic equilibrium in optical depth gives $P^{3/2}=8g[1/2-(3\tau +2{)}^{-1}]/({\kappa }_0{T}_e^8)$. Its logarithmic temperature-pressure gradient is $9\tau /16$, so the Schwarzschild criterion for a monatomic gas is crossed at ${\tau }_c=32/45$. The radiative continuation beyond that depth is unstable and must be replaced by convective transport.

The ionizing photon escape fraction is the fraction of photons capable of photoionization that leave their source environment. Only these escaping photons contribute to ionizing distant intergalactic medium. Its value can depend on frequency, source geometry, and time.