In solid-body rotation the angular velocity is independent of distance from the rotation axis, so tangential speed is proportional to cylindrical radius. A galactic region with this galaxy rotation curve has radial epicyclic frequency $\kappa =2\Omega$.

For $a>b>0$, substitution $z=e^{i\theta }$ gives poles at zero and the two roots of $b{z}^2+2az+b$. The inside-pole residues sum to $(a-\sqrt{a^2-b^2})/(i{b}^2)$, soThe double pole at zero is essential; omitting it gives the wrong sign and value.

For $0<a<1$, integrate $e^{az}/(1+e^z)$ around a rectangle of imaginary height $2\pi$. Its single pole at $\pi i$ has residue $-e^{\pi ia}$, the top-edge integral is $-e^{2\pi ia}$ times the bottom-edge integral, and both vertical sides vanish at large real width. The residue theorem gives ${\int }_{\mathbb{R}}{e}^{ax}/(1+e^x)dx=\pi /\sin (\pi a)$. Substitution $t=e^x$ gives the equivalent integral ${\int }_0^{\infty }{t}^{a-1}/(1+t)dt$.

A subrepresentation is an invariant subspace of a group representation or Lie algebra representation, with the restricted action.

A free massive spin-two field is a symmetric tensor obeying a massive Klein-Gordon equation, vanishing divergence, and vanishing trace. Its rest-frame polarizations form the symmetric traceless square of the $SO(D-1)$ vector, with $(D-2)(D+1)/2$ degrees of freedom.

Microcausality requires local bosonic observables at spacelike separation to commute. For a free real scalar field, the field commutator is the difference of its two Wightman functions; it vanishes at equal time and, by Lorentz invariance, at every spacelike separation. Nonzero spacelike vacuum correlations do not violate this condition, since a correlation is different from the response to a local intervention. Odd fermionic fields satisfy a corresponding spacelike anticommutation condition, while physical even local observables commute.

In an additive mean model, the partial residual for component $j$ subtracts every fitted term except that component. It equals the full regression residual plus the fitted component. Plotting it against that component's predictor shows the conditional pattern after accounting for other fitted effects, and supplies the response for a backfitting update.

A smooth function on a real open interval is real analytic exactly when, near each point, there are constants $C,A>0$ such that $|f^{(n)}(x)|\le C{A}^{n}n!$ for every $n$ and every $x$ in that neighborhood. The bound must hold on a neighborhood, not only at its center. The Taylor theorem with Lagrange remainder proves sufficiency; differentiating a convergent power series on a smaller interval proves necessity. A flat function shows why pointwise bounds alone are insufficient.

A dense viscous crust of thickness $a$ above lighter liquid can overturn on a Rayleigh-Taylor instability time ${\mu }_s/(\Delta \rho ga)$. If diffusive growth gives $a=A\sqrt{\kappa t}$, matching its age to that time gives ${\tau }_s\sim [{\mu }_s/(\Delta \rho gA\sqrt{\kappa }){]}^{2/3}$. A renewed crust then supplies average solid-volume flux $J_s\sim (\Delta \rho g/{\mu }_s{)}^{1/3}(A^2\kappa {)}^{2/3}$. This scale assumes that heat supplied by the interior does not arrest the crust before instability.

A reduced fraction $a/b$ has coprime integers $a,b$ and positive denominator $b$. Distinct reduced fractions with denominators at most $Q$ differ by at least $Q^{-2}$, since their difference has a nonzero integer numerator over a denominator at most $Q^2$.

Replace an estimator by its conditional expectation given selected information. The mean is unchanged and the variance cannot increase, by the law of total variance. In simulation this produces conditional Monte Carlo estimators. Sufficiency is needed for the classical inferential strengthening, but not for the elementary variance reduction identity.

Given a real unit-vector Gram matrix $Y_{ij}=⟨v_i,v_j⟩$, draw a standard Gaussian random vector $Z$ and set $y_i=\mathrm{sign}⟨v_i,Z⟩$. Each coordinate is almost surely a sign; zero projections have probability zero. The same random separating hyperplane is used for every coordinate, so the signs need not be independent. Their pair expectations follow the Gaussian sign-correlation identity.

Colour $n>0$ by $(v_2(n)\mathrm{mod}2,\lfloor {\log }_{2}n\rfloor \mathrm{mod}2)$. No increasing infinite sequence can make all its sums $x_i+x_j$ and $x_i+2x_j$, $i<j$, one colour. If its 2-adic valuations are bounded, two terms of equal valuation and equal odd part modulo four give pair sums of different valuation parity. If they are unbounded, a later term divisible by a sufficiently large power of two prevents adding a fixed earlier term from crossing either dyadic boundary; the two resulting sums have adjacent logarithmic scale indices.

A hydrogen atom state with principal quantum number $n=l+1$ has no radial nodes and radial factor $R(r)=A{r}^l{e}^{-r/[a(l+1)]}$, where $a$ is the Bohr radius. Its energy is $-{\hslash }^2/[2m{a}^2(l+1{)}^2]$. With unit-normalized spherical harmonics, the radial normalization gives $|A{|}^2=2^{2l+3}/[a^{2l+3}(l+1{)}^{2l+3}(2l+2)!]$.

For a normalized maximal-angular-momentum hydrogen state, the probability inside a small radius $r_p$ is the integral of $|R{|}^2{r}^2$. Since $R$ behaves as $r^l$ near zero, it equals $2^{2l+3}(r_p/a{)}^{2l+3}/[(2l+3)!(l+1{)}^{2l+3}]$ times $1+O(r_p/a)$. Increasing angular momentum strongly suppresses penetration toward the Coulomb centre.

This quiver has one vertex and $r$ loops. Paths are words in the loops, so its path algebra is a free associative algebra. A representation assigns an arbitrary endomorphism to each loop.

The quiver has two vertices and one arrow. Splitting a representation's kernel, image and complementary target decomposes it into source-only and sink-only simples and copies of the identity $k\to k$. These are its three indecomposables. Fixed-dimension isomorphism classes are rank orbits of a matrix under left-right multiplication.

This quiver has $r$ source vertices, each with one arrow to a common sink. Its path algebra has the vertex idempotents and the arrows as basis. After extracting source kernels from a representation, its remaining arrows identify the source spaces with subspaces of the sink.

An acyclic quiver has no oriented cycle. For a finite quiver, this is equivalent to finite-dimensionality of its path algebra: paths have bounded length without cycles, while powers of a cycle are infinitely many distinct basis elements.

The Kronecker quiver has two vertices and two parallel arrows from the first to the second. For a representation with arrows $I,N$, an endomorphism consists of the same matrix at both vertices and must commute with $N$. If $N$ is a single Nilpotent Jordan block, this centralizer is $k[N]$, yielding an indecomposable non-brick when the block size exceeds one.