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For an integer period $S$, ${\Delta }_S{X}_t=X_t-X_{t-S}$ uses the backshift operator as $1-B^S$. It annihilates a deterministic period-$S$ mean. If $X_t=m_t+{\epsilon }_t$ with $m_{t+S}=m_t$ and strong white noise, the result is the stationary moving average ${\epsilon }_t-{\epsilon }_{t-S}$.

A scalar matrix is a scalar multiple of the identity matrix, equivalently a diagonal matrix with all diagonal entries equal. It commutes with every square matrix of the same size. In $G{L}_n(F)$, the nonzero scalar matrices form the central subgroup removed to define the projective linear group.

The single-spin density of states counts momentum states for one spin species. For degenerate up and down species, the spin-summed density of states is twice this value. Specify separately whether the density is per unit volume or extensive. In the BCS gap equation, the momentum sum labels opposite-spin pairs once and therefore uses the single-spin density per volume when the interaction has normalization $V/\mathcal{V}$.

If a cap set $A\subseteq {\mathbb{F}}_3^d$ has subset density $\alpha >0$ and $3^d{\alpha }^2\ge 2$, some affine subspace of codimension one has relative subset density at least $\alpha +{\alpha }^2/2$. The zero-sum count and the Parseval identity on a finite group give a nonzero finite abelian Fourier coefficient of magnitude at least ${\alpha }^2/2$. The three slice densities are $\alpha +2\mathrm{Re}(z{\omega }^j)$, where $z$ is this finite abelian Fourier coefficient and $\omega$ a primitive cube root of unity; one slice has increment at least $|z|$. Translation preserves the zero-sum condition because the characteristic is three.

The smoothed matter density variance measures the root-mean-square linear density contrast after applying a window corresponding to a Lagrangian mass scale. For a homogeneous isotropic field it is ${\sigma }^2=(1/(2{\pi }^2))\int k^{2}P(k)|W(kR){|}^{2}dk$. The variance convention, growth epoch and smoothing mass-density relation must agree with the collapse barrier used in a halo abundance calculation.

In matter domination with the constant growing potential, the full Newtonian-gauge energy constraint gives the stated density contrast. The second term dominates inside the Hubble radius and grows as $a$, because ${\mathcal{H}}^2\propto a^{-1}$. Outside the Hubble radius, the constant $-2\Phi$ term dominates and is gauge dependent. The comoving density combination ${\Delta }_c={\delta }_c+3\mathcal{H}{\theta }_c/k^2$ cancels it for ${\theta }_c=2k^2\Phi /(3\mathcal{H})$, leaving the cosmological Poisson equation relation. Matter-spectrum power laws must specify whether they refer to this comoving density or to subhorizon Newtonian-gauge density.

For a null-homologous knot, choose a meridian of a knot $\mu$ and Seifert longitude $\lambda$. Rational surgery with relatively prime numerator $p$ and denominator $q$ fills the knot exterior along $p\mu +q\lambda$. A negative continued fraction for $p/q$ replaces this operation by integer surgery on a chain of meridians, using slam-dunk moves.

Let a knot lie in a fiber of a mapping torus of an oriented topological surface. Surgery with coefficient $-1$ relative to its surface framing replaces the monodromy by its composition with a right-handed Dehn twist. Coefficient $+1$ gives the inverse twist. The composition side depends on the choice of seam and time orientation; the existence of the new fiber bundle does not.

The average vertex degree of a graph is the sum of its vertex degrees divided by its order. Double counting edge endpoints gives $\overline{d}(G)=2e(G)/|V(G)|$. It lies between the minimum degree of a graph and the maximum degree of a graph.

The embedding line bundle has at least four independent global sections. If it is nonspecial, the Riemann-Roch theorem gives $h^0=6-g$, so $g\le 2$. If it is special, the Clifford inequality for curves gives $h^0\le 3$, a contradiction. The smooth projective curve need not span its full ambient projective space.

For two outgoing particles the Lorentz-invariant phase space is $d{\Phi }_2=(2\pi {)}^4{\delta }^{(4)}(p-k-k^{\prime })d^{3}k/((2\pi {)}^{3}2E_k)d^3{k}^{\prime }/((2\pi {)}^{3}2E_{k^{\prime }})$. With massless daughters in their center-of-mass frame, $d{\Phi }_2=d\Omega /(32{\pi }^2)$ and $\int d{\Phi }_2=1/(8\pi )$. This fixes normalization factors in decay widths and scattering rates.

For a pseudoscalar meson of mass $M$ decaying into a pseudoscalar meson of mass $m$ and two massless leptons through a weak charged current, the transverse massless leptonic current selects $f_{+}(s)$. With hadronic current normalized as $(p+k{)}^{\mu }{f}_{+}(s)+(p-k{)}^{\mu }{f}_{-}(s)$ and flavour coefficient $V$,The Källén function has mass dimension four, and the coefficient has mass dimension $-7$. This decay rate assumes the stated form-factor normalization and no additional isospin factor. It follows by the integrated massless leptonic tensor and the pion momentum change of variable in the parent's centre-of-momentum frame.

A lateral sinusoidal surface charge density with wave number $k$ in the Debye–Hückel approximation generates vertical exponential or hyperbolic functions with inverse length $p=\sqrt{k^2+{\kappa }_D^2}$. In a homogeneous electrolyte filling all space, an isolated sheet at $z_s$ produces $\varphi ={\sigma }_k(x)e^{-p|z-z_s|}/(2\epsilon p)$. Boundary conditions determine the normalization when the electrolyte occupies only a gap.

A population of distinct, nonoverlapping dark matter halos cannot contain more total mass per comoving volume than the cosmic mean matter density. A narrow mass population therefore obeys $n_c{M}_h\le {\bar{\rho }}_{m,0}$. Do not sum subhalo masses and their inclusive host masses in the same bound, because that double-counts material. This check can reveal incompatible inferred absorption incidence, object size and halo-mass assumptions.

A Darcy velocity $\mathbf{u}$ driven by pressure in a stationary porous skeleton dissipates mechanical power $\mu |\mathbf{u}{|}^2/\Pi$ per bulk volume. It follows by dotting Darcy law with the discharge: $-\mathbf{u}\cdot \nabla p=\mu |\mathbf{u}{|}^2/\Pi$. Its contribution to a heat equation can be neglected only when it is small relative to the relevant thermal-energy transfer.

For the normalized damped pendulum, the sum of kinetic energy and gravitational potential energy, divided by $mg\ell$, isThe chain rule and the full nonlinear equation give this identity exactly, without a small-angle approximation. Here $\dot{\theta }=d\theta /d\tau$ as in damped pendulum. In physical units the mechanical energy and the power of the resisting torque areThe pendulum's lost mechanical energy is transferred to the damper and surroundings; it does not violate conservation of energy. In an ideal viscous dashpot this loss is frictional heating. The exact cumulative balance is particularly useful for checking a numerical solution:

The figure uses $c=0.20$ and release from rest. Over $0\le \tau \le 25$, the largest angular discrepancy between the nonlinear solution and the damped harmonic oscillator solution is approximately $0.000702\mathrm{rad}$ ($0.0402^{\circ }$) for a $10^{\circ }$ release, but $0.14995\mathrm{rad}$ ($8.592^{\circ }$) for a $60^{\circ }$ release. Their nonlinear initial energies are $1-\cos 10^{\circ }\approx 0.015192$ and $1-\cos 60^{\circ }=0.5$, respectively. At $\tau =25$ the remaining fractions are approximately $0.0064080$ and $0.0063757$. These are illustrative calculations from specified initial conditions, not observations. The generator uses a Runge-Kutta method and integrates $Q$ alongside the motion; halving its step and independently integrating $c{\dot{\theta }}^2$ checks the result.

For $\ell =1\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the first three nonlinear same-side swing intervals after a $60^{\circ }$ release are approximately $2.1008$, $2.0378$, and $2.0221\mathrm{s}$. They approach the linear-model oscillation period $2.0162\mathrm{s}$ as the amplitude shrinks. After a $10^{\circ }$ release the corresponding intervals are already about $2.0185$, $2.0168$, and $2.0164\mathrm{s}$. This makes the experiment's period drift a quantitative test of the small-angle approximation, rather than an automatic reason to change the damping coefficient.

At a turning point, $\dot{\theta }=0$ and $\dot{E}=0$ even though the swing is losing mechanical energy over the cycle. Its potential energy there is $E_n=1-\cos A_n$, so measured turning angles give successive energies without estimating angular velocity. In the proposed decay experiment, multiply these values by $mg\ell$ to obtain energies in joules, then subtract adjacent same-side energies to estimate the loss over each full swing. Only for small amplitudes is $E_n\simeq A_n^2/2$. In the linear underdamped model, same-side peak energies consequently satisfy $E_{n+1}/E_n=e^{-\gamma T_d}$, the square of the angular amplitude ratio. For the illustrative parameters above this energy ratio is approximately $0.2828$. Between turning points, kinetic energy and potential energy continue to exchange while their sum decreases; the energy slope vanishes briefly at each extremum and is negative whenever the bob moves.

It is therefore inaccurate to replace the exact identity by $\dot{E}=-cE$. Even for the linear model, $E_{\mathrm{lin}}=({\dot{\theta }}^2+{\theta }^2)/2$ has ${\dot{E}}_{\mathrm{lin}}=-c{\dot{\theta }}^2$. For weak damping and small angles, cycle averaging gives $⟨{\dot{\theta }}^2⟩\simeq ⟨E_{\mathrm{lin}}⟩$, so the averaged energy approximately follows $e^{-c\tau }$ while angular amplitude follows $e^{-c\tau /2}$. This exponential decay is an approximation to the energy trace; the same-side peak ratio is exact within the linear model. The figure distinguishes the actual energy curves from this averaging approximation.

The energy identity also gives a global conclusion without solving the nonlinear equation. If $E$ were constant on a positive-length interval, then $\dot{\theta }$ would vanish throughout it; the equation would force $\sin \theta =0$, making the entire trajectory an equilibrium. Every nonstationary trajectory therefore loses mechanical energy over every positive-length interval, even though its instantaneous derivative can vanish at a turning point. A nonconstant periodic motion would have to return to its initial mechanical energy, so it is impossible for this undriven damped pendulum. Oscillations near a stable focus are successive, shrinking swings, not a repeating cycle. Moreover, if $E(0)<2$, the angle cannot reach an upright equilibrium, whose potential energy is $2$; such motion stays within its original potential well. This is the usefulness of the energy description: one scalar balance controls both the observed decay and which nonlinear motions remain accessible.

The weak topology of a CW complex is determined by its closed cells: a subset is closed if and only if its inverse image under every characteristic map from a closed cell is closed. A map out of the complex is continuous when its restrictions along those characteristic maps are continuous. This is distinct from the weak topology of a vector space. Taking the product with a locally finite CW complex, such as a closed interval with its finite cell structure, retains the corresponding CW weak topology.

Attaching a $q$-cell along $\varphi :S^{q-1}\to X$ forms the quotient of $X\bigsqcup D^q$ identifying each boundary point with its image under $\varphi$. A constant attaching map gives $X\vee S^q$. An attaching map can affect the cohomology ring even when it does not change the additive cohomology groups.

A CW pair consists of a CW complex $X$ and a subcomplex $A$. Its inclusion is a cofibration, so homotopies on $A$ extend across $X$. The relative lifting property of a Serre fibration for such pairs allows specified boundary lifts to be retained while lifting a homotopy.

For an all-integer dictionary row $x_B+{\sum }_j{a}_j{x}_j=b$ with nonnegative variables, this cut follows because $x_B+{\sum }_j\lfloor a_j\rfloor x_j$ is an integer and ${\sum }_j\{a_j\}{x}_j\ge 0$. Consequently that integer is at most $\lfloor b\rfloor$. Negative coefficients use the usual fractional part $u-\lfloor u\rfloor$, so $\{-1/5\}=4/5$. Normalize a resulting inequality sensibly before introducing a new integer-valued slack variable.