For a down-set of size $m$, maximize its edge boundary by minimizing the sum of the sizes of its members. Choose all sets in the smallest ranks, followed by any needed part of the next rank; this minimizes that sum among all families and is itself a down-set. With $M_j={\sum }_{i=0}^j\left(\genfrac{}{}{0ex}{}{n}{i}\right)$, $M_{-1}=0$, and $M_{r-1}\le m\le M_r$, the exact maximum is $h(m)=nm-2[{\sum }_{j<r}j(\genfrac{}{}{0ex}{}{n}{j})+r(m-M_{r-1})]$.

Linear eccentricity damping on one planet changes $\dot{z}=iAz$ to $\dot{z}=i[A+i\mathrm{diag}({\tau }^{-1},0)]z$. For a nondegenerate, weakly damped two-planet matrix, let $D=\sqrt{(A_{11}-A_{22}{)}^2+4A_{12}{A}_{21}}$. The modal damping rates are $g_{p,m}=[1\pm (A_{11}-A_{22})/D]/(2\tau )>0$. Both secular eigenmodes damp even though only one planet is directly affected; the coupling shares the dissipation.

Every doubly stochastic matrix is a convex combination of permutation matrices. Equivalently, the extreme points of the doubly stochastic convex polytope are exactly the permutation matrices. The extreme-point assertion follows from an alternating-cycle perturbation of a doubly stochastic matrix; the convex-hull assertion follows because a bounded finite-dimensional polytope is the convex hull of its vertices. Permutation matrices are extreme since their zero entries force the same zeros in every convex decomposition.

The total differential is $d_h+d_v$ for anticommuting differentials. For a double cochain complex bounded in both indices, both index filtrations are finite and have the same total cohomology as abutment.

Filtering the total cochain complex by each index yields two spectral sequences. One first takes vertical cohomology and the other horizontal cohomology; boundedness ensures both converge to total cohomology.

The displayed vector calculus identity follows by contracting the two alternating tensors in the component formula for curl. If both fields have zero divergence, only the two directional derivatives remain. It rewrites the transport terms in the adjoint linearized Navier-Stokes evolution.

For a locally integrable vector field, this formula defines its divergence as a distribution. For a piecewise smooth field, a jump of its normal component across an interface adds a surface measure to the smooth divergences. A continuous normal component eliminates that measure, as follows by integration by parts on the two sides.

For a vector field directed radially from a fixed axis and depending only on distance $\rho$ from that axis, cylindrical coordinates give $\nabla \cdot [f(\rho ){\mathbf{e}}_{\rho }]=f\prime +f/\rho$. Zero divergence gives $f=A/\rho$ on any positive radial interval. Unless $A=0$, this field is singular on the axis, so applications of the divergence theorem must use a domain avoiding it or account separately for that singularity.

For $U^{\prime \prime }=(\alpha I-K_h)U$ with symmetric positive definite $K_h$, all-data displacement is bounded for all time on a fixed grid exactly when $\alpha <{\lambda }_{\min }(K_h)$. Diagonalization of a matrix reduces this assertion to oscillators. At equality, an initial velocity in a zero-frequency mode produces linear growth; above it a growing hyperbolic mode is available. A lower bound on ${\lambda }_{\min }(K_h)-\alpha$ uniform in the mesh gives a uniform displacement estimate. This all-time statement is stronger than displacement stability of a symmetric semidiscrete wave equation on fixed finite time intervals.

The critical level of the regular edge wave $c=\Lambda /m$, $U=\Lambda z$, is $z_c=1/m$ for $\Lambda \ne 0$. At that height $U=c$, so mean advection and the phase term oppose in the wave frame. A literal condition $U=-c$ instead gives the unphysical height $-1/m$ below the lower wall.

A disk flares when its disk aspect ratio increases outward. This is stronger than the statement that its height increases: a disk with $H\propto r$ has constant aspect ratio and is not flaring by this definition. The radial temperature and surface density of a disk profiles determine flaring in a specified vertical equilibrium model.

For fixed volume and equal ambient viscosities, radial volume conservation gives $u=r/(5t)$. Summing both ambient-face tractions gives $H^2=32\mu R/(5\pi \Delta \rho gt)$ and $R^5=125\Delta \rho g{V}^{2}t/(512\pi \mu )$. The edge slope is singular, so the profile is a leading bulk thin-current approximation.

For the bounded Dirichlet Poisson integral in a quadrant, the Wirtinger derivative isThis follows by differentiating the Schwarz integral formula composed with $z^2$, remembering that for real boundary data $u_z$ is half the derivative of a holomorphic function whose real part is $u$. For complex boundary data, apply the construction to the real and imaginary parts separately and use linearity. If the data and their derivatives decay sufficiently, integration by parts givesThe endpoint terms cancel exactly when $g_1(0)=g_2(0)$. Interior denominators do not vanish because $\mathrm{Im}{z}^2>0$.

The harmonic function $v(x,y)=xy$ vanishes on both axes but is nonzero in the quadrant, and $v_z=-iz/2$. More generally $\mathrm{Im}{z}^{2m}$, for a positive integer $m$, vanishes on both axes and is harmonic. Adding these unbounded functions leaves all edge Dirichlet boundary conditions unchanged. Thus decay of the boundary data does not imply decay of every solution, and an edge-data-only formula cannot determine an arbitrary solution's Wirtinger derivative. A boundedness or suitable growth condition must select the solution.

Two bounded harmonic functions continuous to the quadrant edges with the same Dirichlet boundary conditions are equal. Their difference is bounded and zero on both axes. Odd reflection across each axis gives a bounded harmonic extension away from the corner; the removable singularity for a bounded harmonic function fills in that corner. The harmonic Liouville theorem makes that extension constant, and oddness makes the constant zero. This does not exclude unbounded examples such as the zero-boundary harmonic growth in a quadrant.

For a particle with mass $m$ on $(0,L)$, put $K_0(x;\beta )=\sqrt{m/(2\pi {\hslash }^2\beta )}{e}^{-m{x}^2/(2{\hslash }^2\beta )}$. The method of images gives $K_D(q_f,q_i;\beta )={\sum }_r[K_0(q_f-q_i+2rL;\beta )-K_0(q_f+q_i+2rL;\beta )]$. The diagonal integral isThe reflected intervals tile the real line, giving the subtraction $1/2$. The Poisson summation formula converts this to ${\sum }_{n\ge 1}{e}^{-\beta {\hslash }^2{\pi }^2{n}^2/(2m{L}^2)}$, proving equality between the image and energy eigenstate calculations.

For a finite irreducible Markov chain with at least two states satisfying detailed balance, its Dirichlet form of a Markov chain obeysExpand $f-\pi (f)$ in an orthonormal eigenbasis. The denominator is ${\sum }_{j\ge 2}{a}_j^2$, and the numerator is ${\sum }_{j\ge 2}(1-{\lambda }_j)a_j^2$. Their ratio is at least $1-{\lambda }_2$ and equality holds for the second eigenfunction. Thus the optimal Poincare inequality for a reversible Markov chain constant is $1/\gamma$, for the ordinary rather than absolute spectral gap.

For a finite extension of number fields, the different exponent satisfies $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}-1$, with equality exactly when the extension at $\mathfrak{P}$ is tamely ramified. The finite residue fields are perfect, so tameness is equivalent to the residue characteristic not dividing the ramification index. Thus unramified primes have exponent zero and wildly ramified primes have exponent at least $e_{\mathfrak{P}}$. No Galois extension hypothesis is needed.

For $K_r={\mathbb{Q}}_p({\zeta }_{p^r})$, the uniformizer ${\zeta }_{p^r}-1$ has different exponent $p^{r-1}(r(p-1)-1)$. Differentiate the cyclotomic polynomial at ${\zeta }_{p^r}$: the numerator contributes $r{p}^{r-1}(p-1)$ to the valuation, while the denominator ${\zeta }_p-1$ contributes $p^{r-1}$.

A dictator is the fixed agent whose top-ranked alternative is always selected by a dictatorship in social choice.