The linear growth condition bounds the drift norm and diffusion-matrix norm by $C(1+|x|)$ for one constant $C$. Together with globally Lipschitz coefficients it gives global strong existence and pathwise uniqueness, preventing finite-time explosion. Linear coefficient maps satisfy it even when they are not bounded.

In an incompressible shearing sheet with background velocity $-2Ax{\mathbf{e}}_y$, take both the perturbation velocity and the entire magnetic field to be real single Fourier modes with a common real wavevector and transverse amplitudes. Then every quadratic directional derivative vanishes, including terms involving complex conjugates. There is no uniform background magnetic field and consequently no linear magnetic-tension coupling between the amplitudes. Advecting the phase requires ${\dot{k}}_x=2A{k}_y$, ${\dot{k}}_y={\dot{k}}_z=0$. The velocity amplitude obeys $\dot{\tilde{\mathbf{v}}}-2A{\tilde{v}}_x{\mathbf{e}}_y+2\mathbf{\Omega }\times \tilde{\mathbf{v}}=-i\mathbf{k}\tilde{\psi }-\nu k^2\tilde{\mathbf{v}}$; the magnetic amplitude obeys the displayed equation. The modified pressure includes magnetic pressure; the physical pressure can contain a mean and second harmonic even when the modified pressure is a single mode.

If a random variable has variance zero and mean $m$, it equals $m$ almost surely. For a target probability strictly between zero and one, $\mathbb{P}(X\ge C)$ meets the target exactly when $m<C$. Substituting zero variance into a positive-variance non-strict quantile constraint incorrectly includes $m=C$, where the overflow probability is one. The distinction comes from the atom at the threshold.

For an $h$-oriented real bundle $\xi$ of rank $r$, pull its Thom class back along the zero section to obtain its Euler class in $h^r(X)$. It depends on the orientation and the actual bundle, not merely its stable isomorphism class. In complex cobordism the orientation may be supplied by a stable complex structure on a real vector bundle.

Take a connected sum of $r$ copies of Complex projective space ${\mathbb{CP}}^3$ and $s$ copies of $S^3\times S^3$, with the empty sum interpreted as $S^6$. Their Euler characteristics are four, zero and two, respectively. Removing a six-ball subtracts one from the Euler characteristic and gluing along its boundary $S^5$ subtracts zero, so each connected sum subtracts two. The result is $2+2r-2s$. For any even integer $n$, choose $r=\max (n/2-1,0)$ and $s=\max (1-n/2,0)$.

If $c$ is a regular element of a ring in a right Noetherian ring and a nonzero right ideal $B$ had $B\cap cR=0$, cancellation would make $B+cB+c^{2}B+\cdots$ an infinite direct sum of nonzero right ideals. Its finite partial sums contradict the ascending chain condition. Thus $cR$ is an essential right ideal.

Even restricting to increasing sequences in $[0,1]$ with limit of a sequence one gives no common stopping index. For any positive integer $K$, defineThis is an increasing bounded sequence converging to one, but its absolute error is one through index $K$. Since $K$ can exceed any proposed index, the bounded monotone sequence theorem alone cannot say when the absolute error will fall below $1/2$. A zero difference between two successive terms is also insufficient: the initial plateau eventually ends.

There is no universal eventual rate either. Let $(r_n)$ be any proposed positive nonincreasing sequence tending to zero, with $r_n\le 1$. SetThen $(a_n)$ is increasing, lies in $[0,1]$, and converges to one, but for every sufficiently large $n$ its absolute error satisfies $1-a_n=\sqrt{r_n}>r_n$. Indeed, $\sqrt{r_n}/r_n=1/\sqrt{r_n}\to \infty$, so no fixed constant multiple of $r_n$ bounds the absolute error eventually. If a proposed rate is not nonincreasing, replace it, after a finite initial segment if necessary, by the decreasing envelope $s_n=\sup \{r_k:k\ge n\}$; this also tends to zero, and the same construction defeats the original rate. This is an original construction, rather than a quantitative claim attributed to the qualitative theorem.

For a concrete slow example, $a_n=1-1/\log (n+2)$ increases to one, where $\log$ is the natural logarithm. Its absolute error is $1/\log (n+2)$, so making it less than $\epsilon$ requires $n+2>\exp (1/\epsilon )$, using the exponential function. At $\epsilon =0.1$, this first happens at $n=22025$. The reciprocal and geometric examples in increasing and decreasing approximations to a common limit meet the same tolerance at $n=11$ and $n=4$, respectively. All three obey the same bounded monotone sequence theorem.

For rational $a$, a multiple of its denominator gives exact integer recurrence. For irrational $a$, Weyl differencing and the Weyl criterion give an equidistributed sequence $a{m}^2$ modulo one. Thus every parameter belongs to one of the open sets $\{a:\Vert a{m}^2\Vert <ϵ\}$. Compactness of the circle group gives a finite subcover, hence a bound $N$ uniform in $a$; no explicit rate follows from this compactness step.

For $|a|\ll \mu$, the equal-density ion-neutral cubic has a rapidly damped mode ${\omega }_d=-2i\mu +i{a}^2/(4\mu )+O(a^4/{\mu }^3)$ and two propagating modes ${\omega }_{\pm }=\pm a/\sqrt{2}-i{a}^2/(8\mu )+O(a^3/{\mu }^2)$. The reduced phase speed uses the total density ${\rho }_i+{\rho }_n=2{\rho }_i$. The exact displacement ratio is ${\xi }_n/{\xi }_i=\mu /(\mu -i\omega )$. It is nearly $-1$ for the rapidly damped counterflow and nearly $1\pm ia/(\sqrt{2}\mu )$ for the propagating modes. Thus the latter have small ion-neutral slippage of relative size $|a|/\mu$, responsible for ambipolar damping.

For a characteristic-p local field with residue field ${\mathbb{F}}_q$, the simple roots of $X^q-X$ lift uniquely by Hensel lemma. Their lifts form a copy of ${\mathbb{F}}_q$, since qth powers preserve addition and multiplication. Choose a uniformizer $\pi$. Repeatedly subtracting the lifted residue and dividing by $\pi$ produces a unique convergent expansion ${\sum }_{i\ge 0}{a}_i{\pi }^i$ for each integral element. Allowing finitely many negative powers gives a valued-field isomorphism $K\cong {\mathbb{F}}_q((T))$, $T\mapsto \pi$.

First prove the Engel lemma by induction on the dimension of a Lie algebra of nilpotent endomorphisms. If $x^m=0$, then $\mathrm{ad}x=L_x-R_x$ on the endomorphism algebra has $(\mathrm{ad}x{)}^{2m-1}=0$, since left and right multiplication commute. Thus a proper Lie subalgebra $M$ acts nilpotently on $L/M$, and induction gives a nonzero class normalized by $M$. This proves the Engel normalizer lemma.

For a maximal proper $M$, its normalizer is all of $L$, so $M$ is an ideal. Maximality then forces $\dim L/M=1$, since every line in a Lie algebra is a subalgebra. Induction gives a nonzero common annihilator $W$ of $M$ in the representation space. Ideality makes $W$ stable under $L$. A generator $x$ of $L/M$ is nilpotent on $W$ and has nonzero kernel there, providing a vector killed by all of $L$. In the possibly nonfaithful quotient actions, apply induction to the image, whose dimension is at most $\dim M$.

Repeatedly apply this common-kernel statement to quotient representation spaces to obtain a complete flag with $x{V}_j\subseteq V_{j-1}$. Hence the operators are simultaneously strictly upper triangular. Applying this to the Adjoint representation shows that a finite-dimensional Lie algebra whose every adjoint map is nilpotent has terminating lower central series. Conversely, termination of that series immediately makes every adjoint map nilpotent. This proves Engel theorem.

// Target: semisimple-lie-algebra.bigb

For dipole pressure $\delta p\sim Cx/(x^2+y^2)$, choose a rectangular control volume whose transverse size tends to infinity faster than its streamwise size. Each end face contributes $\pi C$ to $\int \delta p{n}_{x}ds$, so fixed-throughflow boundary work gives $\Delta \mathcal{D}=-2\pi UC$, the displayed result. This control-surface choice removes the integrated far-field velocity correction. A circular cutoff of the pressure-only functional gives $-\pi UC$ instead and does not implement the same throughflow normalization.

After eliminating a vertex set $E$, a generic edge between remaining vertices $i,j$ exists exactly when the original matrix graph contains a path from $i$ to $j$ whose internal vertices all belong to $E$. Inductively, the new edges at pivot $k$ concatenate two such paths through $k$; conversely split any allowed path at $k$, or retain it when it avoids $k$. This characterizes fill by graph reachability.

An electrical conductor supports mobile charge carriers and hence an electric current under a field. In a band description, a partially filled band or available overlapping bands supply nearby accessible states. The contrast with a band insulator depends on band filling as well as spectral gaps; semiconductors provide carriers through thermal excitation or doping.

A uniaxial electrical conductivity has one distinguished direction $n$ and equal response in its orthogonal plane. Its eigenvalues are $\alpha$ in $n^{\perp }$ and $\alpha +\gamma$ along $n$, so it is invertible precisely when $\alpha (\alpha +\gamma )\ne 0$. For an electric field perpendicular to $n$, the electric current density is $J=\alpha E$. Positive dissipation requires both eigenvalues to be nonnegative, a physical constraint separate from algebraic invertibility. For $D_{ij}={ϵ}_{ijk}{n}_k$, one has $D^2=n{n}^{\mathsf{T}}-I$, and $\sigma D^2=-\sigma$ is equivalent to $\gamma =-\alpha$; the conductivity then vanishes along $n$.

Electrical resistivity is the inverse of scalar electrical conductivity in an isotropic Ohmic medium. In its rest frame, $\mathbf{E}={\rho }_{\mathrm{el}}\mathbf{j}$, so the electric field does work per unit volume $\mathbf{j}\cdot \mathbf{E}={\rho }_{\mathrm{el}}|\mathbf{j}{|}^2$. In Gaussian units, finite conductivity gives magnetic diffusivity $\eta =c^2/(4\pi \sigma )$; neglecting this diffusion is a distinct approximation from neglecting electron inertia in Hall magnetohydrodynamics.

Complex conductivity relates the complex amplitudes of current density and electric field at an angular frequency $\omega$ in sinusoidal response. Its modulus gives the amplitude ratio, while its argument encodes their relative phase. Both the exponential time convention and whether amplitudes are peak or root mean square values must be stated when interpreting phase or average power.

Select the three downgoing or depth-decaying elastic modes for a solid below a plane free surface. For each mode use its reduced traction column $b=(Tq+R)a$. Zero total boundary traction gives $BA=-b_0$ for an incident mode of reduced traction $b_0$. Invertibility gives the displayed amplitudes. A singular matrix instead permits an outgoing homogeneous surface mode, so a radiation-limit or solvability analysis is necessary.

For integer $q\ge 2$, given a random-cluster configuration, its components receive independent uniform colors under the Edwards-Sokal coupling. Two vertices have equal Potts spins with probability $1$ if connected and $1/q$ otherwise. Averaging proves the identity. With a wired component fixed to color $1$, the same argument gives normalized spontaneous magnetization $(q{\pi }^1({\sigma }_0=1)-1)/(q-1)={\varphi }^{\mathrm{w}}(0\leftrightarrow \infty )$ in infinite volume. Thus its onset is the wired random-cluster critical probability, with dimensionless Potts model coupling $K_c=-\log (1-p_c)$ in the convention of the Edwards-Sokal coupling.

For $q=1$, the equal-spin probability is identically one and the first identity reduces to $1=1$. The normalized magnetization expression divides by $q-1$ and therefore does not apply to that degenerate one-color case. The conversion to a physical Ising model inverse temperature must also account for the interaction convention.

The Cornish-Fisher expansion inverts an Edgeworth expansion to approximate quantiles. For a standardized iid sum, let ${\gamma }_1$ be the skewness of one summand and $z_{\alpha }$ the standard normal quantile. The first distribution function correction is $-{\gamma }_1(z^2-1)\varphi (z)/(6\sqrt{n})$. Substituting $y_{\alpha }=z_{\alpha }+c/\sqrt{n}$ and expanding $\Phi$ gives $c={\gamma }_1(z_{\alpha }^2-1)/6$. This is a fixed-quantile asymptotic, with $\alpha$ fixed in $(0,1)$.