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For a compact set $J$ homeomorphic to a circle, lifting an argument along its parametrization identifies the quotient by continuous exponentials with $\mathbb{Z}$. By planar logarithm factorization it is also free abelian on the bounded complementary components, so exactly one exists. A proper closed arc has trivial exponential quotient, and therefore connected complement. Given a small open arc of $J$ inside any neighborhood of a point, join points from the two complementary regions by a path avoiding the remaining closed arc. The first and last intersections with $J$ occur in the small arc, proving that the neighborhood meets both regions. Thus $J$ is the common boundary of its unique bounded and unbounded regions.

Draw a $k\times D$ random matrix $G$ whose entries are independent random variables with the standard normal distribution, and setThe factor $1/\sqrt{k}$ makes the expected value of the squared L2 norm equal to the original squared L2 norm. This Gaussian Johnson–Lindenstrauss transform is a linear map, generally not an orthogonal projection: its rows are not required to be orthogonal vectors. ^[ref]

For the fixed finite set $S$, let $0<\delta <1$ be an allowed failure probability, and defineWith $\log$ denoting the natural logarithm, a sufficient integer target dimension isThen one draw of this specified random matrix satisfies all the squared Euclidean distance bounds simultaneously with probability at least $1-\delta$. Since $c(\epsilon )\ge {\epsilon }^2/12$, the simpler sufficient bound $k\ge \lceil 12{\epsilon }^{-2}\log (n(n-1)/\delta )\rceil$ also works. Both are sufficient bounds for this argument; they do not claim the smallest possible target dimension.

This is a probability statement about a map drawn independently of the fixed finite set. It implies existence because the success probability is positive, but permits failed draws. The set cannot be chosen after inspecting the map with this same guarantee. In particular, no single draw preserves every possible finite set in the ambient Euclidean space when $k<D$.

A centrally symmetric convex body is in John position when its John ellipsoid is the Euclidean unit ball.

For a centrally symmetric convex body in John position, there are contact points $u_j\in \partial K\cap S^{d-1}$ and positive weights $c_j$ such that ${\sum }_j{c}_j{u}_j\otimes u_j=I$. Taking the trace gives ${\sum }_j{c}_j=d$. A supporting hyperplane at $u_j$ shows $|⟨u_j,x⟩|\le \Vert x{\Vert }_K$.

Here is the maximum-volume argument for the decomposition. If $I$ were outside the convex cone generated by the contact matrices, a separating hyperplane would give a symmetric matrix $A$ with $\mathrm{tr}A>0$ and $⟨Au,u⟩\le 0$ at every contact. Subtract a small positive multiple of $I$ so these inequalities become strict while the trace remains positive. The support function of $(I+tA)B_2^d$ is $|(I+tA)v|$. It decreases below $1$ near all contact directions for small positive $t$; away from them, the strict gap $h_K(v)>1$ preserves containment. Its volume increases, since $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$, a contradiction. The cone is closed because its generators have trace $1$ and form a compact set; finite-dimensional convexity gives a finite decomposition.

In inclusive electron-positron annihilation into hadrons, hard gluon radiation produces a quark, an antiquark and a gluon, giving three resolved particle jets. Observation of such events at PETRA in 1979 provided direct evidence for gluons. The institutional account is www.desy.de/news/news_search/index_eng.html?openDirectAnchor=1643&printversion=1 .

Apply Jensen's inequality to the two equally weighted values $a,b$ and the convex function $t\mapsto t^p$. Equality for $a=b>0$ proves that $2^{p-1}$ is the smallest possible constant. For $p>1$, strict convexity makes equality equivalent to $a=b$; for $p=1$, equality holds for every pair.

The Jensen gap is the nonnegative difference $\mathbb{E}[f(X)]-f(\mathbb{E}X)$ in Jensen's inequality, under its integrability assumptions. It quantifies how much applying a convex function before averaging differs from applying it afterward. For $f(x)=x^2$ it is the variance of $X$; for an affine function it is zero. For general convex functions, it depends on both the function and the distribution of $X$, and is not determined by the variance alone.

For a meromorphic function with no zero or pole on $|w|=R$, set $G_R(z,a)=\log |(R^2-\overline{a}z)/(R(z-a))|$. At source points that are neither zeros nor poles, the displayed identity expresses its logarithmic modulus as the Poisson integral of its boundary values, minus its zero Green potentials plus its pole Green potentials. The Poisson kernel is $(R^2-|z{|}^2)/|R{e}^{i\theta }-z{|}^2$ with normalized angular measure. Subtracting the corresponding logarithmic singularities leaves a harmonic function, so the harmonic Poisson formula proves the identity. At zero it reduces to Jensen's formula.

With physical wavelength $2\pi /k$, the Jeans length is $2\pi /k_J$, where the Jeans wavenumber is $\sqrt{4\pi G\rho }/c_s$. Linear pressure support dominates gravity below this length, whereas longer wavelengths can grow. In an expanding background the growth also depends on the Hubble damping term, so the static threshold does not by itself give the time dependence.

For a matter-dominated fluid with constant $c_s^2=w\ll 1$, combine stress-energy conservation in synchronous gauge with the scalar trace Einstein equation. Discard pressure corrections to the background gravitational and velocity-damping terms while retaining the potentially large pressure-gradient term $w{k}^2\delta$. This produces the displayed leading nonrelativistic equation. It is not an exact equation at finite $w$. At negligible pressure the growing and decaying matter-era solutions are ${\tau }^2$ and ${\tau }^{-3}$.

// Target: cosmology.bigb

For increasing containment indicators, attach independent Bernoulli selectors of parameter $t$. Janson inequality applied to the selected events gives $\mathbb{E}(1-t{)}^X\le \exp (-t\mu +t^2\Delta /2)$. Set $t=1-e^{-s}$ and apply the exponential Markov bound. The estimates $s-s^2/2\le 1-e^{-s}\le s$ give exponent at most $-sa+s^2(\mu +\Delta )/2$. Optimizing at $s=a/(\mu +\Delta )$ proves the displayed lower-tail bound, with the dependency sum ordered and excluding the diagonal.

Here $[a]$ is the trivial character of $S_a$, $[0]$ is the induction-product unit, $[a]=0$ for $a<0$, and multiplication is induction of external tensor products. The Frobenius characteristic map turns the expression into the Jacobi–Trudi identity. A sign-reversing involution on intersecting lattice-path systems leaves precisely the nonintersecting systems corresponding to Semistandard Young tableaux, proving the determinant identity.

For independent identically distributed observations with finite fourth moment, the sample mean of their squares has sampling variance $\mathrm{Var}(X^2)/n$. Its jackknife variance estimator is exactly the displayed unbiased estimator, since its leave-one-out values differ from their average by $-(x_i^2-\overline{x^2})/(n-1)$. The conditional bootstrap variance of a sample mean is ${\hat{v}}_B=n^{-2}{\sum }_i(x_i^2-\overline{x^2}{)}^2=(n-1){\hat{v}}_J/n$. Thus the two methods have a known finite-sample scaling difference, despite sharing the same large-sample target.

A reactive strategy cooperates with probability $p$ after the opponent cooperated and probability $q$ after the opponent defected. It ignores its own previous action. Independent behavioural randomization gives a four-state joint-action Markov chain. The reduced marginal recurrences depend on the prescribed initial action distribution.

A grim trigger is a history-dependent pure strategy in a repeated game that switches permanently to punishment after a trigger event. In the iterated prisoner's dilemma above, the trigger is any player's observed defection, and punishment is always defect. Unlike Tit for tat, grim trigger never restores cooperation after the opponent cooperates again. Whether mutual grim trigger is a subgame perfect equilibrium depends on the stage payoffs, discount factor, and monitoring assumptions; it is not an unconditional property of the rule. See Ozdaglar, lecture 15, slides 12–15.

The product of two coordinate iterated integrals of a path is the sum over order-preserving interleavings of their integration variables. Split the product of the two ordered simplices according to the joint order of all variables. This identity explains the algebraic constraints on a geometric rough path and the cancellation of low-degree terms in the composition of its local Euler maps.

For isotropic incompressible flow, directional averaging gives $⟨({\partial }_x{u}_x{)}^3⟩=(8/105)⟨\mathrm{tr}{S}^3⟩=-(2/35)⟨{\omega }_i{\omega }_j{S}_{ij}⟩$. Positive mean vortex stretching therefore implies negative longitudinal derivative skewness. Statistical homogeneity alone does not determine the sign in a fixed direction.

The turbulent energy spectrum $E(k)$ distributes kinetic energy over wavenumber shells, normalized by ${\int }_0^{\infty }E(k)dk=⟨|\mathbf{u}{|}^2⟩/2$. For isotropic incompressible flow its tensor form is ${\Phi }_{ij}=E(k)({\delta }_{ij}-k_i{k}_j/k^2)/(4\pi k^2)$. Its low-wavenumber behavior records large-eddy momentum statistics.

The velocity correlation tensor is $Q_{ij}(\mathbf{r})=⟨u_i(\mathbf{x})u_j(\mathbf{x}+\mathbf{r})⟩$. For isotropic turbulence it splits into longitudinal and transverse parts. Incompressibility constrains them, while the Fourier transform gives the turbulent energy spectrum. Its trace is the scalar velocity-dot-product correlation.

The Saffman integral is $L=\int ⟨\mathbf{u}(\mathbf{x})\cdot \mathbf{u}(\mathbf{x}+\mathbf{r})⟩d^{3}r$. With integrable trace correlation it equals the large-volume limit of $⟨|{\int }_V\mathbf{u}dV{|}^2⟩/|V|$. It is nonnegative and measures momentum-fluctuation density. Under appropriate unforced far-field flux conditions it is invariant.