For $\sqrt{p}|000⟩+\sqrt{1-p}|111⟩$, a projective measurement of the number of ones in one party's $n$ local qubits gives a common type class of size $d_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The conditional pure state is $d_k^{-1/2}{\sum }_{|x|=k}|x,x,x⟩$. It can be converted exactly and deterministically to a uniform common-label state of any dimension $m\le d_k$: one party measures an $m$-element subset $S$ with Kraus operator ${\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)}$, and all parties relabel their common retained labels after receiving $S$. Choosing $m=2^f$ gives $f$ independent GHZ states. Stirling's formula and the weak law of large numbers give $f/n\to h_2(p)$ in probability. Measuring every local bit separately would instead destroy the coherence and leave a product state.

For $x>0$, the decimal significand is the unique $S_{10}(x)\in [1,10)$ such that $x=10^k{S}_{10}(x)$ for an integer $k$. Explicitly, $k=\lfloor {\log }_{10}x\rfloor$ and $S_{10}(x)=10^{\{{\log }_{10}x\}}$, using the floor function and the fractional part. For $0.0042$, $4.2$ and $4200$, the decimal significand is $4.2$ and the first nonzero digit is $\lfloor S_{10}(x)\rfloor =4$. In the classical terminology of logarithm tables, the logarithmic mantissa is $\{{\log }_{10}x\}$, which differs from the decimal significand $10^{\{{\log }_{10}x\}}$.

At $\omega =-3/2$, the combined scalar equation of Brans-Dicke theory becomes the matter consistency condition $T=0$. It does not determine $\square \Phi$ by division, since the wave coefficient has vanished. The original metric tensor and scalar Euler-Lagrange equations must be retained. Introducing scalar-dependent matter or a scalar potential changes the consistency condition.

Choose a rational basis of the dual of a vector bundle and an effective divisor $D$ bounding all its poles. Evaluation gives a generically injective map $E\to {\mathcal{O}}_C(D{)}^{\oplus r}$. A line subbundle maps nontrivially to at least one summand, so ${\mathcal{O}}_C(D)\otimes L^{-1}$ has a nonzero regular section. Consequently $\mathrm{deg}L\le \mathrm{deg}D$. This bound depends on $E$ and the chosen rational frame, and does not require $E$ to be stable.

For a rank-two bundle $E$, the sufficiently positive twist $E(m{p}_0)$ is globally generated. The nowhere-vanishing section of a globally generated bundle on a curve gives a subbundle ${\mathcal{O}}_C\subset E(m{p}_0)$. Twisting back produces a genuine line subbundle of degree $-m$, with locally free quotient. Thus line-subbundle degrees are bounded above but not below. Merely inserting a negative divisor on an algebraic curve into a fixed subline would allow a torsion quotient and would not establish this stronger conclusion.

A sample correlation matrix rescales a sample covariance matrix $S$ to unit diagonal, for variables with positive sample variance. Its off-diagonal entries are sample correlation coefficients. Strong correlations between predictors can indicate multicollinearity, but pairwise correlations alone do not exclude higher-order linear dependence.

If a density has polynomially decaying tails on both half-lines, its moment-generating function is infinite at every nonzero real parameter: exponential growth defeats polynomial decay on the positive tail for a positive parameter, and on the negative tail for a negative parameter. Thus no open real tilting domain exists for a saddlepoint density approximation, even when the distribution has finite mean and variance and a central limit theorem applies.

For a regular model with a suitable ancillary statistic $a$, the displayed expression approximates the conditional density of the maximum-likelihood estimator. The normalizing factor includes $(2\pi {)}^{-p/2}$ for a $p$-dimensional estimator, depending on convention. The observed information is evaluated at the sample's own fitted value. In a pure scale model the factor is proportional to $1/\hat{\theta }$; this Jacobian determinant can make the normalized approximation coincide with the exact conditional density.

For a sample mean from an independent identically distributed sample, let $K(t)$ be the cumulant-generating function of one observation and solve $K^{\prime }(t)=x$. Put $w=\mathrm{sgn}(t)\sqrt{2n(tx-K(t))}$ and $u=t\sqrt{n{K}^{\prime \prime }(t)}$. Under the smooth, nonlattice conditions for a saddlepoint density approximation, the corresponding approximation to the cumulative distribution function iswhere $\Phi$ and $\varphi$ are the standard normal distribution functions. The apparent singularity at $t=0$ is removable. Write $v=K^{\prime \prime }(0)$ and $k=K^{\prime \prime \prime }(0)$; Taylor expansion gives $w=t\sqrt{nv}(1+kt/(3v)+O(t^2))$ and $u=t\sqrt{nv}(1+kt/(2v)+O(t^2))$. Thus $1/w-1/u\to k/(6\sqrt{n}{v}^{3/2})$. The approximation requires suitable tail regularity and is not a universal formula for distributions without a moment-generating function.

For independent unit-scale Laplace distributions, the cumulant-generating function is $K(t)=-\log (1-t^2)$ for $|t|<1$. The saddlepoint density approximation solves $K^{\prime }(t)=y$, giving $t=y/(1+\sqrt{1+y^2})$. At this tilt, $K^{\prime \prime }(t)=D(1+D)$ and $ty=D-1$, giving the displayed probability density function approximation for the sample mean. Its value at zero is $\sqrt{n/(4\pi )}$; it need not integrate to one before normalization.

For fixed $0<A<1$ and $\delta <1$, the two folds of the signed gluing-map reduction enclose three scalar fixed-point branches. Writing $\delta =1-\eta$ gives width $C=\eta (1-\eta {)}^{-1}[A(1-\eta ){]}^{1/\eta }$. Since $(1-\eta {)}^{1/\eta }\to e^{-1}$, the displayed essential asymptotic follows. The fold curves meet with exponentially small width; this is not the ordinary algebraic cusp normal form of a smooth cubic vector field. A negative fixed point lifts to two symmetry-related flow cycles, so the cusp describes distinct return-map branches rather than counting every physical cycle separately.

In a rectangle $[0,m]\times [0,2n]$, $m\ge n$, let $X$ require an open vertical crossing of the corner square $[0,n{]}^2$ attached to the rectangle's right side. Explore the leftmost vertical crossing of the square without inspecting edges to its right. Reflect its geometry in the line $y=n$, obtaining a deterministic separator. Every horizontal rectangle crossing reaches this separator from the right; reflection divides these possible attachments into two equally probable choices, so the lower-half attachment has probability at least half the horizontal crossing probability. The required edges remain unexposed and independent of the explored crossing. Averaging proves the estimate. Reflected geometry need not be open.

The row space of a matrix is the linear span of its rows. It is the column space of the matrix transpose.

For constant electron density, common path length $L$, and a solenoidal isotropic spectral tensor, the rotation-measure correlation contains the window $|{\int }_0^L{e}^{i{k}_{z}z}dz{|}^2$. As $L$ becomes large compared with the correlation length, this window acts as $2\pi L\delta (k_z)$ within the integral. The transverse projector then has $zz$ component one, yielding the displayed two-dimensional Fourier transform. For finite $L$, the windowed integral retains longitudinal modes and endpoint corrections. The method of using such rotation measure correlations to infer magnetic spectra is developed by Enßlin and Vogt.

List each pair $(i,j)$ with multiplicity $a_{ij}$, sorted first by $i$ and then by $j$. Apply row insertion to the lower letters $j$, and record the upper letters $i$ in the new cells. The row insertion comparison is strictly greater than the carried entry. The result is a pair of Semistandard Young tableaux of the same shape. Reverse insertion removes the rightmost cell containing the largest recording entry, then replaces the rightmost strictly smaller entry in each preceding row. It reconstructs the ordered list, proving the bijection. The insertion and recording contents are respectively the column sums and row sums of $A$.

With $ϵ={\beta }^{\prime }/\beta \ll 1$, $\eta ={\mu }^{\prime }/\mu =O({ϵ}^2)$ and a fixed mode index, put $y=qh$, $W=\omega h/{\beta }^{\prime }$ and $K=kh$. The exact Love wave branch has $W=y\sqrt{1+{\eta }^2{\tan }^{2}y}/\sqrt{1-{ϵ}^2}$ and $K=y\sqrt{{ϵ}^2+{\eta }^2{\tan }^{2}y}/\sqrt{1-{ϵ}^2}$. Its minimum group velocity occurs near $y=(n+1/2)\pi$, with transition width $W-(n+1/2)\pi$ of order $({\eta }^2(n+1/2)\pi {)}^{1/3}$. Well above that width, the rigid-base approximation to Love waves captures both velocities. Density ratios and the mode index must be controlled in this estimate.

A càdlàg function is right-continuous and has a finite left limit at every positive time. The name abbreviates the French phrase /continue à droite, limites à gauche/.

On continuous piecewise smooth perpendicular fields with zero endpoints, the Riemannian index form is nonnegative exactly when no point strictly inside the geodesic segment is conjugate to its initial point. It is positive on every nonzero field exactly when no conjugate point occurs up to and including the final endpoint. At a first conjugate endpoint its nullspace consists of the endpoint-vanishing Jacobi fields. These statements distinguish nonnegative quadratic forms from strictly positive ones and do not assume global minimization.

On continuous piecewise smooth vector field along a curve with zero endpoints, the radical of a bilinear form of the Riemannian index form is exactly the space of endpoint-vanishing Jacobi fields. Integration by parts against tests supported inside each smooth piece gives the Jacobi equation; tests with arbitrary values at breakpoints force continuity of the covariant derivative, hence a global smooth Jacobi field. This characterization uses $I(V,W)=0$ for every test field $W$, not merely $I(V,V)=0$, which is insufficient for an indefinite form.

// Target: riemannian-geometry.bigb

In a parallel orthonormal frame along a geodesic, write the Jacobi equation as $A^{\prime \prime }+KA=0$, with $A(0)=0$, $A^{\prime }(0)=I$ and symmetric $K$. In the absence of conjugate points on $(0,b]$, $A$ is invertible there. The conserved quantity $A^T{A}^{\prime }-A^{\prime T}A=0$ makes $S=A^{\prime }{A}^{-1}$ symmetric, and $S^{\prime }+S^2+K=0$. For a field $u$ vanishing at both endpoints,Indeed the difference of the integrands is $(u^{T}Su{)}^{\prime }$; its boundary value at zero vanishes because $S=t^{-1}I+O(t)$ and $u=O(t)$. Equality forces $u=Ac$ and then $A(b)c=0$, so $u=0$. This proves strict positivity and the minimizing property of the Jacobi field with prescribed endpoint values.

// Target: riemannian-geometry.bigb