For backward differences $\Delta P_i=P_i-P_{i-1}$, a scalar stationary arity-$a$ subdivision mask has symbol $M(z)={\sum }_i{M}_i{z}^i$. The identity $(1-z)M(z)=B(z)(1-z^a)$ gives the displayed intertwining whenever the quotient is a finite Laurent polynomial. Iterating supplies higher-difference masks. The coefficients $a^{r\ell }{\Delta }^r{P}_i^{\ell }$ are candidate $r$th-derivative controls. A convergent derivative scheme, proved by contraction of its first differences, establishes corresponding smoothness of the original limit under the standard consistent parameterization.

The subdivision arity is the integer factor $a>1$ by which a stationary refinement shrinks parameter-grid intervals: control index $i$ has coordinate $i/a^{\ell }$ after $\ell$ levels. It is distinct from the mask width and from the number of neighbors in a stencil.

For a stationary univariate refinement $S_{ij}=a_{i-mj}$, the subdivision mask width counts fine-grid positions from the first to the last active entry in a column, including any internal zeros. Consecutive columns are translates by $m$ rows, the subdivision arity. The number of old control points used by a particular new point is instead the size of its row stencil; it need not equal the mask width. A span convention counts $w-1$ fine intervals, so numerical width claims should specify the convention.

For an oriented plane function $\ell$, certified extrema over a bounding volume allow a subdivision curve piece to be rejected when $\max \ell <0$, or wholly retained when $\min \ell \ge 0$. Otherwise subdivide. At an explicitly chosen rendering tolerance, a sufficiently flat, sufficiently short piece can be replaced by its clipped chord. This is an approximation, not a proof of exact crossing topology. Exact treatment of tangent contacts and small excursions requires root isolation of $\ell (C(t))$ and splitting into sign-constant intervals, when the curve representation supports such isolation.

For a point light at $L$, the shadow of a curve point is the first surface intersection on its ray beyond that point. Certified patch bounds and subdivision isolate intersections; regular hits are corrected with a three-variable Newton method solve. Differentiating the displayed equation gives $[S_u,S_v,-(C-L)](u^{\prime },v^{\prime },{\lambda }^{\prime }{)}^T=\lambda C^{\prime }$, a predictor for continuation along the shadow. Surface boundaries and grazing rays require clipping and reseeding; every visible component must be isolated, rather than assumed detectable from a fixed sample of the casting curve.

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 $\nu >0$ statistical degrees of freedom. Equivalently, conditional on $\Lambda =\lambda >0$, $T$ has normal distribution $N(0,\nu /\lambda )$. This is a normal scale mixture representation of Student t: a small latent denominator occasionally produces a much larger spread. The result is a heavy-tailed distribution even though every conditional probability distribution is a normal distribution.

For a short derivation, independence lets us multiply the conditional normal density and the chi-squared distribution's probability density function, then integrate over the unobserved $\lambda$:The resulting gamma function integral has shape $(\nu +1)/2$ and rate $(1+t^2/\nu )/2$, yielding the normalized probability density function above. Also, for $\nu >2$, conditional expectation gives $\mathbb{E}[T^2]=\nu \mathbb{E}[{\Lambda }^{-1}]=\nu /(\nu -2)$. Thus unit scale does not mean unit variance. Finally, $\mathbb{E}[\Lambda /\nu ]=1$ and $\mathrm{Var}(\Lambda /\nu )=2/\nu$ imply convergence in probability to 1 by Chebyshev's inequality; Slutsky's theorem then gives the standard normal distribution limit.

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 $c(G⊠G)=c(G{)}^2$.

For a finite-dimensional convex optimization problem with affine equalities $Ax=b$ and convex inequalities $g_i(x)\le 0$, Slater's condition requires a point in the relative interior of the common effective domain that satisfies the equalities and makes every inequality strict. A refined version only requires strictness of the nonaffine inequalities; affine inequalities may hold with equality. If the primal optimal value is finite, these conditions imply strong duality and dual attainment.^[ref]

The relative-domain hypothesis matters. Minimize $-\sqrt{x}$ over its domain $x\ge 0$ subject to $x\le 0$. The only feasible point is zero, giving primal value zero. For multiplier $\lambda >0$, the dual function isIts supremum is zero but is not attained by a finite multiplier. The feasible point lies outside the relative interior $(0,\infty )$ of the objective domain, so the refined condition does not apply. Feasibility of affine constraints alone suffices only when the relevant domain hypothesis is also met.

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.

Let $K\subseteq {\mathbb{R}}^2$ be compact and convex, and assume that it contains a point with both coordinates positive. The product $xy$ has a unique maximizer over $K\cap [0,\infty {)}^2$, and that maximizer lies in the positive quadrant. Compactness gives existence; the strictly positive attainable product excludes points on the axes. On the positive quadrant, maximizing $xy$ is equivalent to maximizing $\log x+\log y$, a strictly concave function. A segment between two distinct maximizers would contradict strict concavity, proving uniqueness.

The quadrant restriction matters. The line segment from $(-10,-10)$ to $(1,1)$ is a compact convex set meeting the positive quadrant, but its global product maximum over the entire segment is $100$ at the negative endpoint. The result concerns the positive feasible portion, unless the original feasible set is itself confined to the nonnegative quadrant.