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In $ϵy^{\prime \prime }+xy+y^2=0$, balancing all three terms where the outer branches $0$ and $-x$ meet gives the displayed dominant balance and the inner equation $Y^{\prime \prime }+XY+Y^2=0$. Matching to $-X$ and writing $Y=-X+u$ gives $u^{\prime \prime }-Xu+u^2=0$. Its linearized decaying correction is an Airy function; the growing correction is excluded. Matching to zero instead gives $Y^{\prime \prime }+XY=0$, whose nonzero Airy function combinations oscillate. Thus the acceptable matching branches differ qualitatively at the turning endpoint.

For a nonexplosive continuous-time Markov chain on the nonnegative integers, suppose zero is absorbing and each positive state has positive rate to zero. On every finite positive set the conditional chance of absorption at the next jump has a positive minimum. By the Strong Markov property, the chance of surviving $r$ departures from that set is bounded by $(1-\delta {)}^r$. Without absorption only finitely many such departures occur, and finite positive holding rates prohibit staying forever at a positive state. Thus eventually the path stays outside each finite set: it escapes to infinity. This conclusion needs no upward skip-free assumption.

If an escaping stellar wind carries specific orbital angular momentum $\lambda J/M$ with constant $\lambda$, then $\dot{J}/J=\lambda \dot{M}/M$ and $J\propto M^{\lambda }$. Combining this with circular-binary orbital angular momentum gives the displayed period invariant. It is a different prescription from donor-wind angular-momentum loss, whose specific angular momentum generally varies with mass ratio.

Let $x=\log (R/R_L)$, $q=M_1/M_2$, ${\dot{M}}_2=-f{\dot{M}}_1=M_{1}x/t_{\mathrm{dyn}}$, and $d\log R/dt=n{\dot{M}}_1/M_1+t_{\mathrm{nuc}}^{-1}$. Under the binary period invariant for a constant wind angular-momentum factor and $R_L=0.46a(M_1/M{)}^{1/3}$, the restoring coefficient is $A(q)=n+5/3-2fq-2(3\lambda +1)(1-f)q/[3(1+q)]$. Locally freezing the masses gives an attracting overfill $x_{\ast }=f{t}_{\mathrm{dyn}}/[A(q)t_{\mathrm{nuc}}]$ precisely when $A(q)>0$. If $A$ is of order one and $t_{\mathrm{dyn}}\ll t_{\mathrm{nuc}}$, transfer is driven slowly by nuclear expansion. A negative $A$ gives positive feedback on the short overfill timescale; detailed outcomes then exceed this linear contact model.

The range of a finite-dimensional, finite, countably additive, atomless vector measure is compact and convex. In particular, an atomless measure restricted to a measurable set $E$ can realize every mass between zero and $\mu (E)$. This allows exact balancing by small disjoint consumer cohorts in an exchange economy.

For $1<p\le q<\infty$ and $0\le \rho \le \sqrt{(p-1)/(q-1)}$, the noise operator on the Boolean hypercube satisfies $\Vert T_{\rho }f{\Vert }_q\le \Vert f{\Vert }_p$ for the uniform probability measure. In particular,Here $\hat{f}(S)$ are Fourier-Walsh transform coefficients. The endpoint $\rho =0$ follows by averaging. This hypercontractive inequality on the Boolean hypercube controls Fourier-Walsh transform coefficients of discrete derivatives of a Boolean function and is a principal ingredient in the Kahn-Kalai-Linial theorem.

The Noether–Lefschetz theorem makes any curve class a positive multiple of the hyperplane class, with positive self-intersection. But the smooth rational-curve self-intersection on a projective hypersurface formula makes every smooth rational curve have self-intersection at most $-2$ when the surface degree is at least four. Thus none exists. The smoothness condition is essential; this does not prohibit singular curves whose normalizations are rational.

Assume a column is sealed to salt at both ends, diffusion occurs only in liquid, and its imposed temperature increases strictly upward. A finite-width equilibrium mushy layer cannot be stationary: its solute flux is nonzero unless its liquid fraction vanishes. The final state is pure ice below a height $h_f$ and uniform-composition brine above it. Their interface obeys $C_f=-(T_E+G{h}_f)/m$. The displayed mass conservation equation fixes $h_f$ from the initial salt content. For a diffusive steadily pulled equilibrium mush, let $H=-T_E/G$, $c=m{C}_0/G$, $d=D/V$ and $h_0=H-c-d$. The initial salt content is $N=C_0[h_0+d\log (1+h_0/c)]$. Thus the liquid-cap depth ${\ell }_f=h_0-h_f$ obeys ${\ell }_f({\ell }_f+c+d)=c[h_0+d\log (1+h_0/c)]$, which has one positive root. This conclusion assumes no subsequent exchange with an underlying eutectic reservoir.

A finite-dimensional Lie algebra is nilpotent if and only if every adjoint map ${\mathrm{ad}}_x$ is nilpotent. A Lie algebra of nilpotent endomorphisms can be simultaneously represented by strictly upper triangular matrices.

Each term is the preimage of the center of a Lie algebra after quotienting by the preceding term. Equivalently, $Z_{i+1}=\{x:[\mathfrak{g},x]\subseteq Z_i\}$. A Lie algebra is nilpotent exactly when this series reaches the whole algebra after finitely many steps. For the Heisenberg Lie algebra, the series is $0\subset \mathbb{C}c\subset \mathfrak{g}$.

Every Adjoint representation operator raises the lower central series of a Lie algebra filtration. In a basis adapted to that finite filtration, all adjoint operators are strictly triangular. Their products have zero trace, so the Killing form vanishes. Consequently its radical of the Killing form is the whole algebra. For strictly upper triangular matrices the filtration can be seen directly by the gap between the column and row indices.

// Target: algebra.bigb

For a vector field with linear part $\dot{u}=v$, $\dot{v}=0$ and cubic terms $a_i{u}^3+b_i{u}^{2}v+c_{i}u{v}^2+d_i{v}^3$, write $(u,v)=(x,y)+(h_1,h_2)$ with homogeneous cubic $h_i$. The transformed cubic field is $f_3+Ah-DhA(x,y{)}^T$, where $A(x,y{)}^T=(y,0{)}^T$. Taking the coefficients of $h_2$ to be $(-a_1,c_2/2,d_2,0)$ and those of $h_1$ to be $((b_1+c_2/2)/3,(c_1+d_2)/2,d_1,0)$ eliminates every cubic term in $\dot{x}$ and the $x{y}^2,y^3$ terms in $\dot{y}$. The remaining normal form is $\dot{x}=y$, $\dot{y}=P{x}^3+Q{x}^{2}y$ through cubic order. The invariant damping coefficient contains the contribution $3a_1$ from differentiating the first equation.

A newform is a normalized Hecke eigenform in the new subspace of cusp forms. Its weight, level and Dirichlet character are part of its specification. For a newform, its Fourier coefficients equal its eigenvalues of Hecke operators.

For a unit nematic director, the bulk elastic density is $[K_1(\nabla \cdot \mathbf{n}{)}^2+K_2(\mathbf{n}\cdot \nabla \times \mathbf{n}{)}^2+K_3|\mathbf{n}\times \nabla \times \mathbf{n}{|}^2]/2$. The three positive moduli penalize nematic splay, nematic twist, and nematic bend. Boundary elastic terms, chirality, or varying nematic order require additional contributions.

For $n$ independent one-stop negative multinomial distribution observations, the natural estimates are $\hat{\alpha }=S_X/(n+S_X+S_Y)$ and $\hat{\beta }=S_Y/(n+S_X+S_Y)$. Their sum is $g((S_X+S_Y)/n)$ with strictly concave $g(z)=z/(1+z)$, so Jensen's inequality makes its expectation strictly below $\alpha +\beta$. Conditional category proportions split this bias between both estimates. With zero observed counts of a category, that estimate lies at zero; in a parameter space requiring strict positivity the likelihood supremum is not attained.

If an effective divisor $D$ is linearly equivalent to a negative curve $C$, then $D=C$. Otherwise $D\cdot C=C^2<0$ forces $C$ to be a component of $D$, since distinct irreducible curves have nonnegative intersection multiplicities. Write $D=C+D^{\prime }$ with $D^{\prime }$ effective and linearly equivalent to zero. An ample divisor intersects $D^{\prime }$ positively unless $D^{\prime }=0$, while linear equivalence forces that intersection to vanish.

Distinct negative curves cannot have the same integral cohomology class: their mutual intersection would equal a negative self-intersection, contradicting nonnegative intersection multiplicities for distinct curves. A smooth complex projective surface has finitely generated integral cohomology groups, so its negative curves form an at most countable set. Zero-self-intersection curves need not obey this restriction: one ruling of ${\mathbb{P}}^1\times {\mathbb{P}}^1$ gives an uncountable family.

Complete the square as $bF+\xi b^2/2=\xi (b+F/\xi {)}^2/2-F^2/(2\xi )$. Translation of the Gaussian functional integral gives the displayed identity for nonzero $\xi$, with the appropriate oscillatory contour prescription. At $\xi =0$, the Nakanishi-Lautrup field is a Lagrange multiplier imposing the gauge-condition delta functional. This field has no kinetic term and no independent propagating degree of freedom.

For the unit-width symmetric box kernel for density estimation, suppose the design density is positive and differentiable at $x$, the regression function is twice differentiable there, and the conditional variance and design density are bounded near $x$. Write $G_n=(nh{)}^{-1}{\sum }_i(Y_i-m(x))1_{\{|X_i-x|\le h/2\}}$. Symmetry cancels its linear Taylor term, giving $\mathbb{E}{G}_n=O(h^2)$; independence gives $\mathrm{Var}{G}_n=O((nh{)}^{-1})$. The absolute error bound for a ratio estimator with a controlled denominator gives the displayed rate if the small-denominator contribution is negligible. The balance $h\asymp n^{-1/5}$ gives $O(n^{-2/5})$. Interior design positivity matters: support boundaries can prevent the linear cancellation.

A Graeco-Latin square is the superposition of two orthogonal Latin squares of the same order. Each symbol of either square occurs once per row and column, and every cross-square pair occurs once. It permits four mutually orthogonal additive factor spaces: rows, columns, and the two symbol factors.