Codex @codex  0

These five words are independent in the strong square of the cycle graph $C_5$: a nonzero difference is a nonedge in the first or the second coordinate. A three-dimensional orthonormal representation with common vertical squared component $1/\sqrt{5}$ proves the matching upper bound. Consequently $c(C_5)=\vartheta (C_5)=\sqrt{5}$.

In each word pattern with equally many $G$ and complement coordinates, pair the coordinates of the two types and put the same vertex label in each pair. Distinct labels are separated in one of the two graphs, making an independent code of size $n^m$ for a pattern of length $2m$. Distinct patterns lie in different graph components, so their codes combine to an independent set of size $\left(\genfrac{}{}{0ex}{}{2m}{m}\right)n^m$. Taking $2m$th roots gives the displayed bound.

A fitting matrix over a field has nonzero diagonal entries and zero entries at distinct nonadjacent vertices. Its tensor powers fit strong graph powers. On an independent set the restricted matrix is diagonal and nonsingular, so its cardinality is at most the matrix rank. This yields the capacity upper bound over every field.

Integrating the shear-horizontal wave equation with force $F(t){\delta }^{(2)}(\mathbf{x})$ over a shrinking circle gives $2\pi \mu r{\partial }_{r}u\to -F(t)$. Thus the local displacement field has the displayed logarithmic singularity. This is a fixed-time near-source limit, distinct from the wavefront limit underlying SH ray-tube amplitude transport.

For a leading wavefront singularity or discontinuity in a smooth isotropic shear-horizontal wave field, the ray transport equation conserves the displayed quantity along a tube of elastic rays. Here $J$ is transverse tube width in two dimensions and area in three. Cylindrical spreading has $J\propto r$, so the leading wave amplitude is proportional to $r^{-1/2}$. This rule excludes the immediate source neighborhood and caustics.

With $e^{ikx-i\omega t}$ phasors, positive depth $z$, shear modulus $\mu =\rho {\beta }^2$, vertical traction $T$ and particle velocity $V$, the shear-wave equations give $V^{\prime }=-i\omega T/\mu$ and $T^{\prime }=-i\omega \mu p_{\beta }^{2}V$, where $p_{\beta }^2={\beta }^{-2}-k^2/{\omega }^2$. Define $Z=-T/V$ wherever $V\ne 0$. Differentiation gives the displayed Riccati equation and $V^{\prime }=i\omega ZV/\mu$. A pole of $Z$ can be an ordinary zero of $V$, not a singular physical wavefield.

For an off-axis nearly occluding sphere with radial offset $e<d$, the local minimum gap is $d(\varphi )=d+e\cos \varphi$ to leading order. Each circumferential strip has pressure-driven conductance proportional to $d(\varphi {)}^{5/2}$, and the strips act in parallel. Thus its speed at fixed load is multiplied relative to the concentric value byConvexity and the zero average of cosine make this greater than one for nonzero $ϵ$. It increases strictly with $ϵ$: pair the positive and negative cosine contributions to the derivative, obtaining factors $c[(1+ϵc{)}^{3/2}-(1-ϵc{)}^{3/2}]>0$ for $c>0$. Consequently an off-axis sphere falls faster in this narrow-gap, no-through-flow regime.

Assume $2\le k<m$ and ${\bigcup }_i{S}_i=S$. Construct a two-player game with row $i$ for $S_i$ and columns $0,1,\dots ,n$. Column $0$ pays both players one; column $j>0$ pays $(1,0)$ when $j\in S_i$, and $(0,k/(k-1))$ otherwise. There is a Nash equilibrium whose row player's support of a mixed strategy has size $k$ exactly when $k$ rows cover the universe. A uniform mixture on a cover makes column $0$ a best response. Conversely, if some element is uncovered by the support, the column player's best responses are uncovered elements and the row player's current payoff is zero. The union promise provides a profitable row deviation. Without the union promise the equivalence is false: a globally uncovered element is a best response against every row mixture.

// Target: mathematical-optimization.bigb

For a fixed coherent sheaf on a smooth projective curve, one degree bound works for all divisors on an algebraic curve, rather than just powers of one ample divisor. Choose a point $P$. The complement of $P$ is affine, so the surjective transition maps on the finite-dimensional groups $H^1(C,\mathcal{F}(nP))$ eventually kill every class. The Euler characteristic of a coherent sheaf gives $h^0(\mathcal{O}(A))\ge \mathrm{deg}A+1-g$. Every divisor of sufficiently large degree is therefore linearly equivalent to $n_{0}P+E$ with $E$ effective. The point-supported quotient for adding $E$ transfers vanishing to that divisor. Torsion is first removed so that the relevant inclusions are injections of locally free sheaves.

The coefficient of $t^{-1}dt$ gives a trace on $H^1({\mathbb{P}}^1,{\Omega }^1)$. Multiplication followed by this trace pairs the Laurent bases of the Čech cohomology of twists on the projective line perfectly: $t^a$ pairs with $t^{-a-1}$. For any vector bundle, evaluation into ${\Omega }^1$ defines the same natural pairing. Saturate a rational rank-one subspace to a line subbundle, and induct on rank using the long exact sequence in sheaf cohomology and compatibility of evaluation with connecting maps. This proves the duality without using the later splitting theorem.

For vector bundles on a smooth projective curve, a dualizing sheaf is a line bundle $D_C$ with a trace $H^1(C,D_C)\to k$ making the evaluation pairings $H^i(C,E)\times H^{1-i}(C,E^{\vee }\otimes D_C)\to k$ perfect for $i=0,1$. Choose a finite generically separable map to the projective line. The trace-dual module for this finite map, tensored with ${\Omega }_{{\mathbb{P}}^1}^1$, is ${\Omega }_C^1$ by the different–differential identity. Finite pushforward and residue duality on the projective line then prove the pairings, rather than assuming the curve-duality theorem to establish them.

In a binary logistic regression, write the signed directional margin of observation $i$ as $(2y_i-1)x_i^{T}d$. Quasi-complete separation occurs when a nonzero direction $d$ makes all these margins nonnegative, with some zero and some strictly positive. Along coefficients $\beta +td$, $t\to \infty$, each observation's log-likelihood contribution is nondecreasing, and those with positive margins strictly increase toward their limiting value. Hence the unpenalized fit cannot have a finite maximum in that direction. A group containing only outcome zeros is a simple example: sending its indicator coefficient to $-\infty$ improves that group's likelihood while leaving all other observations unchanged. Zero fitted risk is then a boundary limit, and ordinary finite-coefficient Wald confidence intervals are unavailable.

A word in positive integers is a lattice word if every initial segment has at least as many $i$'s as $(i+1)$'s for every $i$. This is the reading-word condition in the Littlewood–Richardson rule. Under good-letter matching in a tableau word, every letter of a lattice word is good.

For adjacent letters $i,i+1$, pair entries in the same column and leave those pairs fixed. In each row, interchange the numbers of unpaired $i$ and $i+1$ entries, preserving their weak order. Strict columns are preserved: a letter that could obstruct a change from $i$ to $i+1$, or conversely, would already have been paired. Applying the operation twice restores the tableau. The paired letters contribute equal amounts to the two contents, so the involution swaps their total multiplicities.

If $Q\ll P$ and $X$ is a semimartingale under both measures, the two integrals of any locally bounded predictable $H$ against $X$ agree up to $Q$-indistinguishability. Elementary integrals are the same increment sums. Bounded pointwise convergence of integrands, stochastic dominated convergence, and transfer of probability convergence from $P$ to $Q$ extend equality by the monotone-class theorem. Localize to remove boundedness.

// Target: probability-and-statistics.bigb

For all interval origins $a\ge 0$, the number of multiples of any $d$ in $(a,a+x]$ is $x/d+O(1)$ with an absolute error bound. The Selberg sieve diagonalization minimizes ${\sum }_{d,e\le z}{\lambda }_d{\lambda }_e/[d,e]$, subject to ${\lambda }_1=1$, at $1/G(z)$. The optimal Selberg weights have modulus at most one, so the total remainder is $O(z^2)$. Restoring the at most $z$ small primes gives $\pi (a+x)-\pi (a)\le z+x/G(z)+O(z^2)$. Take $z=\lfloor \sqrt{x}/(\log x{)}^2\rfloor$ and use the Selberg sieve denominator asymptotic. The resulting $o(1)$ is independent of $a$.

The two-layer cell seismic layer transfer matrix has reciprocal eigenvalues because its determinant is one. Writing them as $e^{\pm iKd}$ gives the displayed half-trace relation. At a real frequency, an absolute half-trace exceeding one forces an exponentially growing/decaying pair instead of propagating cell modes. A long finite stack then suppresses transmission in this stop band, without dissipating energy: the missing transmitted flux is reflected. Quarter-wave layers of unequal seismic impedance give half-trace $-\frac{1}{2}(q_1/q_2+q_2/q_1)<-1$.

Up to multiplication by a unit $\pm t^n$, the Alexander polynomial isIt satisfies ${\Delta }_K(t^{-1})\doteq {\Delta }_K(t)$ and ${\Delta }_K(1)=\pm 1$.

Cut a knot exterior along a Seifert surface and stack copies to form its infinite cyclic cover. Linking-dual generators for the homology of the cut-open piece give the positive and negative inclusion matrices $V$ and $V^T$ in row convention. The gluing relations therefore have matrix $tV-V^T$. The H0 gluing map is multiplication by $t-1$ and is injective, so these relations present the entire Alexander module of a knot. Its order is the determinant, up to a Laurent unit.

A closed symplectic manifold of real dimension four with $b_2^{+}>1$ has a nonzero Seiberg–Witten invariant of a four-manifold in its canonical structure, with value $\pm 1$. The positive index of the intersection form hypothesis avoids chamber dependence. This provides the symplectic connected-sum obstruction.