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Aliasing is the inability to distinguish different continuous frequencies from sampled values. The identity above says that frequencies separated by an integer multiple of the sampling rate $f_s$ give identical samples of a complex Fourier mode. For real cosines, changing the sign of a frequency also gives the same waveform, explaining the $2$-versus-$8$ example at rate $10$. The Nyquist–Shannon sampling theorem prevents this ambiguity within its specified class of bandlimited functions. Before sampling, unwanted higher frequencies must be suppressed if they would violate that assumption; once two waveforms give the same samples, those samples cannot identify which was present.

Integrate the five Lagrange interpolation polynomials at the equally spaced nodes to obtain the displayed rule. It is exact for polynomials through degree four, and symmetry also makes it exact in degree five. Symmetric endpoint, inner-node and midpoint weights $a,b,c$ satisfy $2a+2b+c=2$, $2a+b/2=2/3$, $2a+b/8=2/5$, giving $a=7/45$, $b=32/45$, $c=12/45$.

If $M,N$ are strictly positive traded numeraires and $M/N$ is a mean-one martingale after normalizing its initial value under $Q^N$, this ratio defines $Q^M$. Conditional change of measure then makes every $Q^N$-martingale asset price $S/N$ into the corresponding $Q^M$-martingale $S/M$, when the required integrability holds. In particular a martingale exchange rate under one currency's bank-account measure has a reciprocal martingale under the other currency's measure when their interest rates coincide.

A quadratic number field is a number field of degree of a field extension two, necessarily of the form $\mathbb{Q}(\sqrt{d})$ for a square-free integer $d\ne 0,1$. Its two field embeddings send $a+b\sqrt{d}$ to $a\pm b\sqrt{d}$. If $d>0$, both images lie in the real numbers, giving a totally real number field; if $d<0$, they form a pair related by complex conjugation, giving an imaginary quadratic field. The ring of integers of a quadratic field depends on $d$ modulo four.^[ref]

A number field is totally imaginary if it has no embedding into the real numbers. Its embeddings into the complex numbers occur entirely in conjugate pairs. This is also called totally complex; it does not assert that every element of the field is nonreal.

Column density is the number density integrated along a line of sight. Multiplication by an absorption cross-section gives an optical depth. A small column can result from a small local density, a small path length, or a small fraction of atoms in the absorbing state, so density inference requires these quantities to be constrained.

In the ingoing Eddington-Finkelstein coordinates metric $d{s}^2=-f(r)d{v}^2+2dvdr+r^{2}d{\Omega }^2$, where $f=1-2M/r$, the tangent $U=-E{\partial }_r$ is affine when $E>0$ is constant along each ray. The displayed radial partner obeys $N^2=0$ and $U\cdot N=-1$. Since ${\Gamma }^v{}_{rv}={\Gamma }^a{}_{rr}=0$ and ${\Gamma }^r{}_{rv}=-f^{\prime }/2$, its derivative ${\partial }_r{N}^r=f^{\prime }/(2E)$ cancels the connection term, giving ${\nabla }_{U}N=0$. Its screen is the angular two-plane, with optical tensor ${\hat{B}}_{AB}=-Er{\gamma }_{AB}$, null expansion $-2E/r$, and zero null shear and null twist.

For affine null tangent $U$, choose an auxiliary null vector $N$ with $U\cdot N=-1$ and parallel transport it along the rays. The screen-space projector is $P_{ab}=g_{ab}+U_a{N}_b+N_a{U}_b$. The optical matrix of a null congruence describes evolution of screen-projected deviation vectors. Its trace is the null expansion, its symmetric trace-free part is the null shear, and its antisymmetric part is the null twist. In four dimensions $\hat{B}=\theta P/2+\hat{\sigma }+\hat{\omega }$.

The classical stress-energy tensor of a canonical scalar field has a gradient-product term and a metric-proportional term. Contracting with a real null vector removes the latter, leaving the displayed square for a real field. This proves the null energy condition for a minimally coupled canonical scalar, including the massless case. It is not a statement about arbitrary curvature couplings, wrong-sign kinetic terms or a renormalized quantum stress-energy tensor.

For $T_{ab}=(\rho +p)u_a{u}_b+p{g}_{ab}$ and nonzero null $k$, contraction gives $T_{ab}{k}^a{k}^b=(\rho +p)(u\cdot k{)}^2$. A timelike fluid velocity has nonzero scalar product with every nonzero null vector, so the null energy condition is exactly the displayed inequality. Individual nonnegativity of pressure or density is unnecessary for null focusing. A cosmological constant term has zero null contraction.

The volume ratio of a $d$-dimensional normed vector space is $\mathrm{vr}(E)=(\mathrm{vol}(B_E)/\mathrm{vol}(\mathcal{E}){)}^{1/d}$, where $\mathcal{E}$ is its John ellipsoid. It is unchanged by invertible linear changes of coordinates. In John position, polar integration gives $\mathrm{vr}(E{)}^d={\int }_{S^{d-1}}\Vert u{\Vert }_E^{-d}d\sigma (u)$, with normalized spherical measure. Consequently $\sigma \{\Vert u{\Vert }_E\le t\}\le (\mathrm{vr}(E)t{)}^d$.

Given $f$ in a normed vector space and a vector subspace $U$, an element $p_{\ast }\in U$ is a best approximation if $\Vert f-p_{\ast }\Vert ={\inf }_{p\in U}\Vert f-p\Vert$. Existence is automatic for a finite-dimensional vector subspace: a minimizing sequence can be bounded by the triangle inequality, and a subsequence converges by compactness in finite dimension. An arbitrary infinite-dimensional vector subspace need not contain a minimizer. The norm matters: best uniform approximation uses the supremum norm, while $L^1$ approximation minimizes integrated absolute error.

A normed vector space is strictly convex when distinct unit vectors have their midpoint strictly inside the unit ball. The Lp space is strictly convex for $1<p<\infty$: strict convexity of $|z{|}^p$ gives a strict integrated midpoint inequality for distinct unit functions. The endpoint spaces $L^1$ and $L^{\infty }$ need not have this property.

A normed vector space is uniformly convex if, for each $\epsilon >0$, there is $\delta >0$ such that unit-ball vectors $x,y$ with $\Vert x-y\Vert \ge \epsilon$ satisfy $\Vert (x+y)/2\Vert \le 1-\delta$. Completeness is not part of this definition; a complete example is a uniformly convex Banach space.

In a nonzero normed vector space, a unit vector $x$ is approached from outside the unit ball by $(1+\epsilon )x$, so no boundary point is interior. It is approached from inside by $(1-\epsilon )x$, so every boundary point is in the closure of the open ball. The zero space satisfies the same identities trivially. These statements need the radial structure of a normed space; a two-point discrete metric space with interpoint distance one disproves both analogous claims for general metric balls.

In a cyclic group of odd prime order, the normalized ordered three-term arithmetic progression count is ${\sum }_r\widehat{1_A}(r{)}^2\widehat{1_A}(-2r)$. The zero frequency contributes ${\alpha }^3$. If the count is less than ${\alpha }^3/2$, its nonzero-frequency contribution has magnitude greater than ${\alpha }^3/2$. The Parseval identity on a finite group bounds that magnitude by $\alpha {\max }_{r\ne 0}|\widehat{1_A-\alpha }(r)|$, proving the displayed estimate. The count includes progressions of zero common difference.

If the normalization of an integral scheme is flat, then it is an isomorphism. On an affine chart, let $A\subseteq B\subseteq \mathrm{Frac}(A)$ be the integral closure. Integrality gives surjectivity on spectra, so flatness makes $B$ faithfully flat. For $b=u/v\in B$ with $u,v\in A$ and $v\ne 0$, faithful flatness gives $vB\cap A=vA$. Thus $u=va$ for some $a\in A$, and $b=a$. Consequently $B=A$. This proof does not need finiteness of the normalization.

For an algebraically closed field $k$ and a finitely generated field extension $K/k$ of transcendence degree one, choose $t$ with $K/k(t)$ finite. The integral closures of the two standard coordinate rings of the projective line in $K$ are finite modules and glue to a finite normalization of an algebraic curve of the projective line. A one-dimensional normal local ring is a discrete valuation ring; over the perfect field $k$ the resulting projective curve is smooth. Extension of rational maps gives uniqueness of this smooth projective model.

On $x_3^2=x_1^2+x_2^2$ with $x_3>0$, the gradient of the defining function gives a normal proportional to $(-x_1,-x_2,x_3)$. Its Euclidean norm is $\sqrt{2}{x}_3$, proving the displayed normalization. This upward normal lies on the boundary of the Lorentz cone, so self-duality of a second-order cone makes its pairing with every point of the closed upper cone nonnegative.

The core of a subgroup $H\le G$ isIt is the largest normal subgroup of $G$ contained in $H$. If $H$ has finite index $k$, the action on $G/H$ gives $[G:{\mathrm{Core}}_G(H)]\le k!$.