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At a regular differentiable interior nearest point of a parametric curve, the displacement from the query point is perpendicular to the tangent vector. A branch and bound search uses distances to certified enclosing bounding volumes as lower bounds, and evaluated limit points as upper bounds. Refine the most promising restricted pieces until the upper bound differs from the global lower bound by the chosen distance tolerance. Safeguarded Newton root-finding iteration accelerates local candidates, but endpoints, nonregular points, ties and other stationary branches must not be omitted.

A finite subcover is a subcover with finitely many members, including possibly none when $K$ is the empty set. Requiring a finite subcover for every open cover is the definition of compactness.

For an independent sample from a normal distribution, maximization of the likelihood over the unknown variance gives the likelihood ratio for a fixed mean against an unrestricted mean as displayed, where $t=\sqrt{n}(\bar{X}-{\mu }_0)/S$ and $S^2={\sum }_i(X_i-\bar{X}{)}^2/(n-1)$. It decreases with $|t|$. Under the null, the normal sample mean is independent of its centered sum of squares, so $t$ has the exact Student t-distribution with $n-1$ degrees of freedom. The resulting two-sided test requires no asymptotic chi-squared approximation.

For independent normal samples of sizes $m,n$ with a common unknown variance and equal means under the null, $Z=(\overline{X}-\overline{Y})/(\sigma \sqrt{1/m+1/n})$ is standard normal. The independent centered sample sums of squares give $V=(m+n-2)S_p^2/{\sigma }^2\sim {\chi }_{m+n-2}^2$, independently of $Z$. Thus $Z/\sqrt{V/(m+n-2)}$ has the exact Student t-distribution, giving the statistic $(\overline{X}-\overline{Y})/(S_p\sqrt{1/m+1/n})$. A lower one-sided test uses its lower-tail probability.

For independent $Z\sim N(0,1)$ and $\Lambda \sim {\chi }_{\nu }^2$, the displayed ratio has the Student t-distribution with $\nu$ degrees of freedom. Equivalently, $T\mid \Lambda \sim N(0,\nu /\Lambda )$. Integrating the normal conditional density against the chi-squared distribution gives a density proportional to $(1+t^2/\nu {)}^{-(\nu +1)/2}$. A latent scale therefore supplies heavy tails while retaining convenient normal conditional calculations. For $\nu >2$, its variance is $\nu /(\nu -2)$, so its scale parameter is not its standard deviation.

Take four distinct objects $a,b,x,d$ and six nonidentity arrows $m:a\to b$, $h:x\to b$, $r,s:b\rightrightarrows d$, $k:a\to d$, $l:x\to d$, with $rm=sm=k$ and $rh=sh=l$. These and identities are all morphisms. The only nonidentity epimorphisms have target $d$, which admits no arrow to $b$, so every lifting square against $m$ either has an identity left side or does not exist. Thus $m$ is a strong monomorphism. It is not a regular monomorphism: $h$ equalizes $r,s$ but cannot factor through $m$, and an identical pair has $1_b$ as equalizer. Every strong nonregular monomorphism in any category occurs as the image of $m$ under a faithful functor from this category. Indeed, such a monomorphism is not epic, so choose distinct $r,s$ agreeing on it; failure of its equalizer property then supplies $h$.

Let $X$ switch between $\pm 1$ at rate $q$ and be killed into zero at rate $\lambda$. To exceed a new level it must cross it in state $+1$. Restarting there by the Strong Markov property gives the multiplicative identity ${\psi }_{+}(c+d)={\psi }_{+}(c){\psi }_{+}(d)$, hence an exponential tail. First-step conditioning gives ${\psi }_{-}(c)={\int }_0^{\infty }q{e}^{-(q+\lambda )u}{\psi }_{+}(c+u)du$ and ${\psi }_{+}(c)=e^{-(q+\lambda )c}+{\int }_0^{c}q{e}^{-(q+\lambda )u}{\psi }_{-}(c-u)du$. Substituting the exponential determines its rate as $\sqrt{\lambda (2q+\lambda )}$.

For a singular strong limit cardinal $\lambda$, let $\mu =\mathrm{cf}(\lambda )$ and choose cofinal cardinals ${\lambda }_i<\lambda$. Subset restrictions inject $\mathcal{P}(\lambda )$ into ${\prod }_{i<\mu }\mathcal{P}({\lambda }_i)$, whose cardinal is at most ${\lambda }^{\mu }$ by the strong-limit property. Conversely ${\lambda }^{\mu }\le (2^{\lambda }{)}^{\mu }=2^{\lambda }$. This equality alone does not imply the singular cardinals hypothesis.

The capacity is the exponential growth rate of the independence numbers of strong graph powers. Products of independent sets make these numbers supermultiplicative, and Fekete lemma gives the limit and its equality to the supremum. Passing to every second power gives $c(G⊠G)=c(G{)}^2$.

The Slater condition requires a feasible point at which every nonlinear convex inequality is strict and every affine equality holds. For a convex problem it implies strong duality and attainment of the dual optimum under standard finiteness assumptions.

For a closed string in a string sigma model in an isotropic plane wave, insert $e^{-i{\omega }_n\tau +2\pi in\sigma /L}$ into the transverse wave equation. Periodicity requires integer $n$, giving the displayed dispersion relation. The zero spatial mode is an oscillator of frequency $|m\kappa |$; when this mass vanishes it becomes a free particle mode and needs the position-plus-linear-time expansion.

Expand every transverse closed string coordinate in a real orthonormal Fourier basis of the spatial circle. The canonical momenta are $p_{Ia}=T{\dot{q}}_{Ia}$, and spatial orthogonality reduces the field Hamiltonian to the displayed independent harmonic oscillators. Canonical quantization gives ${\omega }_a(a_{Ia}^{†}{a}_{Ia}+1/2)$ for each nonzero-frequency mode; a zero-frequency mode retains the free particle term $p_{Ia}^2/(2T)$. The formal zero-point energy requires a regulator; it is not automatically the massless flat-background intercept. Opposite travelling-wave oscillator combinations diagonalize the translation generator and satisfy closed-string level matching.

The binary Brier score applies squared-error loss to a reported success probability $p$ and the observed indicator $y\in \{0,1\}$. It is a loss, so smaller is better. Its values lie in $[0,1]$: predicting the outcome with certainty gives zero, while assigning it zero probability gives one. Unlike the logarithmic scoring rule loss, the binary Brier score remains bounded even for a confidently wrong forecast.

For a Bernoulli distribution with true success probability $q$, averaging the two squared losses givesThe report controls only $(p-q{)}^2$; the remaining term is the variance of the Bernoulli random variable. Consequently the Brier score is a strictly proper scoring rule. At $q=0.7$, the expected loss is $(p-0.7{)}^2+0.21$, with unique minimum $0.21$ at $p=0.7$. Reporting $p=0.2$ instead gives expected loss $0.46$, despite its better loss on the particular outcome $Y=0$ above.

For several possible categories, the Brier score sums the squared differences between reported probabilities and observed category indicators. With two categories that sum is $2(p-y{)}^2$, so its numerical values are twice the binary convention used here, with the same optimizing report. Brier's original ^[ref] averages these category-sum losses over forecasts. Strictly Proper Scoring Rules, Prediction, and Estimation by Tilmann Gneiting and others (Section 3.1, Example 1, and Table 1) gives the category-sum and binary forms as negative rewards; negating them gives the losses described here.

Reflexification turns a strict partial order into a partial order. Reflexivity is immediate, transitivity follows by the four disjunction cases, and antisymmetry follows because two opposed strict comparisons imply a forbidden self-comparison. The construction uses only coherent logic, so is preserved by inverse image functors of geometric morphisms.

A compact convex subset of ${\mathbb{R}}^2$ meeting the positive quadrant has a unique positive point maximizing $xy$. Existence follows by maximizing the continuous product on the intersection with the closed nonnegative quadrant; a positive attainable value excludes its axes. Uniqueness follows from strict concavity of $\log x+\log y$ on the positive quadrant. This is a useful way to optimize a product even though the product itself is not concave.

If the otherwise force-free second sphere is held against rotation, its angular velocity relative to the incident fluid is $-3a({\mathbf{U}}_0\times \mathbf{n})/(4R^2)$. The rotlet reflected back to the first sphere addsThis follows by evaluating the rotating sphere in Stokes flow at displacement $-R\mathbf{n}$ and using the vector triple-product identity. The holding couple does no work because the held sphere's angular velocity is zero.

Within each prespecified stratum, choose a block containing fixed equal counts of the two treatment assignments, then randomly permute their order. Every completed block gives exact treatment balance. Variable block sizes and allocation concealment protect against predicting the next assignment. Incomplete final blocks permit a small unavoidable imbalance; odd stratum sizes cannot have exact equal allocation.

For constant nonzero Coriolis parameter and positive constant buoyancy frequency, the anisotropic quasi-geostrophic potential vorticity operator becomes the ordinary three-dimensional Laplacian in the displayed coordinates. Its decaying whole-space inverse is $\psi (\mathbf{X})=-\int [Q({\mathbf{X}}^{\prime })-f]/(4\pi |\mathbf{X}-{\mathbf{X}}^{\prime }|)d^3{X}^{\prime }$. Boundary or far-field data are needed to exclude arbitrary harmonic additions.

The Prandtl ratio compares horizontal and vertical scales in rotating density stratification. In the three-dimensional quasi-geostrophic potential vorticity operator, horizontal curvature scales as $\psi /L^2$ and vertical stretching as $(f_0^2/N^2)\psi /H^2$. Comparable terms therefore require $H/L\sim |f_0|/N$. For constant $N$, the stretched coordinate $Z=(N/|f_0|)z$ makes the operator isotropic. This aspect ratio concerns rotational and stratification effects, not the viscous-to-thermal diffusivity Prandtl number.

The compact space $L$ in a stratified neighbourhood ${\mathbb{R}}^s\times cL$ is the link of the stratum. Its dimension is one less than the codimension of that stratum. A disconnected link records distinct local branches.