For known good probability $p={\sin }^2\theta$, choose $j=\lceil \pi /(4\theta )-1/2\rceil$. A known reversible preparation of one state with good probability $p_{\ast }$ permits $j$ ordinary amplitude amplification iterations to reach the good subspace exactly. With ancillary marking access, prepare a flag with one-probability $q=p_{\ast }/p$ and define joint success as good data and flag one. Reflection about the prepared product state is known, while the joint phase test requires a controlled marking oracle or a computed Boolean output. This changes the success projector and enlarges the state space; it does not evade the obstruction to uniform probability lowering by a unitary. Query costs of preparation, its inverse and marking must all be counted.

For $f,g:\kappa \to \kappa$, write $f{<}^{\ast }g$ when there is $\gamma <\kappa$ such that $f(\delta )<g(\delta )$ whenever $\gamma <\delta <\kappa$. For an infinite regular cardinal, suprema of fewer than $\kappa$ ordinals below $\kappa$ remain below it, permitting the long chain under eventual domination construction.

An Euler circuit, also called an Euler tour, is an Eulerian trail whose final vertex equals its initial vertex. Its length is $|E|$. It can revisit other vertices, unlike a Hamiltonian cycle, which visits each vertex once before returning to the start. For a finite undirected multigraph with at least one edge, an Euler circuit exists exactly when the edge support of a graph is a connected graph and every vertex degree is even.

For example, let $p,q$ be distinct parallel edges joining $A,B$, let $\ell$ be a self-loop at $A$, and add an isolated vertex $Z$. Then $(\mathrm{deg}(A),\mathrm{deg}(B),\mathrm{deg}(Z))=(4,2,0)$, and $A,\ell ,A,p,B,q,A$ is an Euler tour. The self-loop supplies one traversal and two incidences; $Z$ need not be visited. Replacing $p,q$ by a single edge changes both vertex degrees to odd numbers and removes the possibility of an Euler circuit. Thus collapsing parallel edges can change the answer.

Two triangles in a graph sharing one vertex $A$ give another instructive Euler circuit: $A,B,C,A,D,E,A$. The shared vertex has vertex degree four, the others have vertex degree two, and $A$ is revisited between the two closed trails. This is precisely the splice used by Hierholzer's algorithm.

An instanton is a localized finite-action classical solution in Euclidean spacetime, used in a semiclassical description of quantum tunnelling and other nonperturbative effects. In one-dimensional quantum mechanics it can connect distinct degenerate potential minima over infinite Euclidean time. A quartic double-well instanton is an elementary example; a Yang-Mills instanton is a gauge-field example with different geometric structure.

For the skew matrix $S=\left(\begin{array}{cccc}0& 1& 1& 1\\ -1& 0& -1& 1\\ -1& 1& 0& -1\\ -1& -1& 1& 0\end{array}\right)$, both Euclidean lattices reduce to the tetracode modulo $3$. A reflection depending on the reduction coset matches their vectors coordinatewise in absolute value. Their weighted length spectra agree for all positive diagonal axis weights. Rationally independent weights force any possible linear isometry to be a coordinate sign change; the code then forces an overall sign, which does not identify the two lattices.

For independent clusters with mean vector $m_i(\theta )$ and positive definite working covariance matrix $V_i$, a generalized estimating equation solvesCorrect mean specification and regularity can give a consistent estimator of identifiable mean coefficients despite a misspecified working covariance. A sandwich covariance matrix accounts for actual cluster variation. Mean equations alone cannot identify two parameters that always enter through the same combination.

For a nonzero elliptic function with period lattice $\Lambda$, its zeros of a function $a_j$ and poles $b_j$, repeated with their multiplicities, have equal total number and satisfy ${\sum }_j{a}_j-{\sum }_j{b}_j\in \Lambda$. Choose a fundamental parallelogram avoiding all zeros and poles on its boundary. The argument principle applied to $f^{\prime }/f$ gives equality of the numbers. Applying the residue theorem to $z{f}^{\prime }/f$ gives their difference of sums. Pairing opposite edges, the extra factors are the two lattice generators multiplied by integrals of $f^{\prime }/f$ along an edge. These integrals are integer multiples of $2\pi i$, because the endpoints have the same nonzero function value. Thus the difference of sums is a lattice element. This links zeros of a pulled-back line to the chord-and-tangent group law.

A power object $PA$ represents parameterized subobjects of $A$: $\mathcal{E}(X,PA)\cong \mathrm{Sub}(X\times A)$. In an elementary topos, it is ${\Omega }^A$, with the membership subobject supplied by evaluation at the subobject classifier.

A logical functor between elementary toposes preserves finite limits, exponential objects and the subobject classifier, with their canonical comparison maps. It therefore preserves power objects and the double-power-object monad. If it has a left adjoint, power-object monadicity and the adjoint lifting theorem for monad algebra functors give it a right adjoint as well.

A Boolean topos is an elementary topos in which every subobject has a complement, equivalently its internal truth-value Heyting algebra satisfies excluded middle. Its internal first-order logic is classical. Being Boolean alone does not imply that it has enough set-valued points.

A noninitial object is irreducible, or supercompact, when every family of subobjects with union the whole object contains the whole object as one member. This concerns unions of subobjects rather than disjoint coproduct decompositions. Irreducible representable sheaves for the regular coverage follow from the one-arrow nature of that coverage.

For the elastic filament operator $K=A{\partial }_x^4$, two integration by parts operations give boundary form $A[\overline{u}{v}^{\prime \prime \prime }-\overline{u^{\prime }}{v}^{\prime \prime }+\overline{u^{\prime \prime }}{v}^{\prime }-\overline{u^{\prime \prime \prime }}v]$. Four elementary homogeneous choices at each endpoint are $h=h^{\prime }=0$ (clamped), $h=h^{\prime \prime }=0$ (hinged), $h^{\prime \prime }=h^{\prime \prime \prime }=0$ (free), and $h^{\prime }=h^{\prime \prime \prime }=0$ (slope-constrained and transverse-force-free). Each defines a self-adjoint operator on a finite interval when imposed at both ends. More general real Robin boundary conditions also give self-adjointness; the four choices are not an exhaustive classification of all boundary subspaces.

A normal mode of a uniformly bending elastic filament fixed in position and slope at both ends satisfies $W^{\prime \prime \prime \prime }=k^{4}W$. With $\beta =kL>0$, the allowed wavenumbers solve $\cos \beta \cosh \beta =1$. The first root is $4.730040745$ and ${\beta }_n=(n+\frac{1}{2})\pi +O(e^{-(n+1/2)\pi })$ for $n\ge 1$. The apparent root at zero is not an eigenfunction: the zero-eigenvalue cubic polynomial satisfying all four clamped boundary conditions is identically zero. This spectrum differs from that of a filament clamped only at one end.

For a positive quadratic bending operator $K$ with a real orthonormal basis of eigenfunctions, the equipartition theorem gives $C(x,y)=k_{B}T{\sum }_n{W}_n(x)W_n(y)/{\mu }_n$, where $K{W}_n={\mu }_n{W}_n$. This is the inverse-operator Green function multiplied by Boltzmann constant and temperature. Unconstrained zero-energy filament modes prevent a normalizable canonical ensemble, and negative modes signal an unstable quadratic model.

For both ends clamped and $K=A{\partial }_x^4$, the diagonal is $C(x,x)=k_{B}T{x}^3(L-x{)}^3/(3A{L}^3)$. This differs from a clamped-free tip variance. The center variance is $k_{B}T{L}^3/(192A)$.

Adding $\frac{1}{2}\int \sigma (x)(h^{\prime }{)}^{2}dx$ to the quadratic bending energy of an elastic filament contributes $-(\sigma h^{\prime }{)}^{\prime }$ to its variational derivative. The endpoint term is $A{h}^{\prime \prime }{\eta }^{\prime }+(\sigma h^{\prime }-A{h}^{\prime \prime \prime })\eta$. If the real filament tension vanishes at both ends, the usual self-adjoint endpoint conditions for filament bending remain valid. A complete eigenfunction expansion diagonalizes the energy even when the eigenfunctions have no explicit formula. Thermal equipartition theorem arguments require positive eigenvalues after fixing any kernel; strong compression can violate this requirement.

A timelike geodesic with unit specific Killing energy can plunge from infinity into a Schwarzschild black hole only if its specific angular momentum satisfies $|h|<4M$. The limiting case $|h|=4M$ asymptotes to the unstable orbit at $r=4M$. A geodesic already inside the angular-momentum barrier can cross the event horizon with larger $|h|$, so the origin-at-infinity condition matters.

For a down-set on $N$ coordinates, let $p_m=|{\mathcal{F}}_m|/(\genfrac{}{}{0ex}{}{N}{m})$. For $1\le m<N$ with $p_m>0$, the effective ground-set parameter of a hereditary uniform layer gives the displayed ratio bound. Iteration is valid up to the first empty layer. In particular, whenever $2m\le N$, $p_{2m}\le p_m^2$; if the later layer is empty this conclusion is immediate. Ratios with $p_m=0$ are undefined and must be replaced by the assertion that all later layers are empty.

For positive dense-fluid reduced gravity $G$ and upward coordinate $z$, an unstable gradient $G_z>0$ supplies buoyant work of order $d^2{G}_z$ over mixing length $d$. velocity $u\sim d\sqrt{G_z}$ and eddy diffusivity $K\sim ud$ follow. Order-one constants depend on the turbulence model. This closure is for an unstable gradient; its square root must not be applied to a stable negative gradient.

A syntheme partitions six points into three duads, equivalently a perfect matching of the complete graph. There are $6!/(2^{3}3!)=15$ synthemes. Every duad belongs to three. Two edge-disjoint synthemes extend to a unique total of synthemes: their union is a six-cycle, whose complement has a unique factorization into three matchings.

For a finite nonempty subset of an abelian group with doubling constant at most $K$, the Plünnecke-Ruzsa inequality gives $|3A-2A|\le K^5|A|$. Applying the Ruzsa covering lemma to $2(A-A)$ using $A$ gives $2T\subseteq D+T$ for $T=A-A$ and $|D|=q\le K^5$. Thus $\ell T\subseteq (\ell -1)D+T$, and counting multiplicities in the finite set $D$ proves the displayed polynomial bound. For fixed $K>1$, it is eventually at most $K^{ϵ\ell }|A|$ for every $ϵ>0$.