For a nonnegative supermartingale and $0<a<b$, let $U$ count completed upcrossings of $[a,b]$, allowing the first entry at time zero. The displayed bound holds for every integer $k\ge 1$. Construct the multiplicative upcrossing supermartingale, stop at the $k$th completed crossing, and use its initial expected value and nonnegativity. The bound is exponential in the number of crossings, and complements the first-moment Doob upcrossing inequality.

If $h>0$ is smooth and $Z_t=h(X_t,t)$ is the unit-mean density martingale of a locally absolutely continuous probability measure relative to Wiener measure, then $h_t+\frac{1}{2}{h}_{xx}=0$. The Itô formula gives $Z=\mathcal{E}(\int (h_x/h)(X_s,s)d{X}_s)$ when $Z_0=1$. The true-martingale hypothesis is already supplied by the density process. On each finite horizon the Girsanov theorem makes $X-\int (h_x/h)(X_s,s)ds$ Brownian under the new measure. Positivity provides local equivalence; global infinite-horizon equivalence and the Novikov condition are not additional requirements.

For a particle far inside a planet's stellar orbit, the planet-centred disturbing function is $G{M}_{\star }/|{\mathbf{r}}_{\star }-\mathbf{r}|-G{M}_{\star }\mathbf{r}\cdot {\mathbf{r}}_{\star }/r_{\star }^3$. Expanding the first term cancels its dipole against the indirect disturbing function. Dropping the spatially constant monopole leaves $G{M}_{\star }{r}^2(3{\cos }^2\psi -1)/(2r_{\star }^3)$, where $\psi$ is the angle between the two offset vectors. The omitted instantaneous terms are $O(G{M}_{\star }{r}^3/r_{\star }^4)$. The gradient of the retained expression is the leading differential stellar tidal acceleration.

Vertices are Lagrangian subspaces of a $2m$-dimensional symplectic vector space, joined when they intersect in codimension one. A pair with intersection dimension $m-r$ admits a symplectic basis in which $U=⟨e_1,\dots ,e_m⟩$ and $W=⟨f_1,\dots ,f_r,e_{r+1},\dots ,e_m⟩$. Replacing $e_i$ successively by $f_i$ gives an isotropic path of length $r$, while one edge can increase intersection dimension with a fixed target by at most one. Equal-distance ordered pairs have the same adapted form, so the symplectic group acts distance-transitively.

A subgroup of the graph automorphism group acts distance-transitively when it is transitive on the ordered pairs of vertices at each fixed graph distance. Thus two vertex pairs at the same distance can be matched by one automorphism from that subgroup. Existence of such a subgroup implies that the full automorphism group has the same property, making the graph a distance-transitive graph. The specified subgroup need not be the entire automorphism group.

The set distance is the infimum of pairwise metric distances. Two nonempty disjoint compact sets have positive set distance: continuity attains the minimum on their Cartesian product, and a zero minimum would give a common point. Disjoint closed sets without compactness need not have positive separation.

For a closed densely defined dissipative operator, the estimate $\Vert (\alpha I-Z)f\Vert \ge \alpha \Vert f\Vert$ makes its range closed and its kernel trivial. Thus a positive spectral value means this range is proper, which is equivalent to a nonzero vector in $\ker (\alpha I-Z^{\ast })$. Dissipativity is essential: a real multiplication operator may have positive continuous spectrum with no adjoint eigenvector.

// Target: analysis.bigb

The Kolmogorov length, velocity and time are $\eta =({\nu }^3/ϵ{)}^{1/4}$, $v_{\eta }=(\nu ϵ{)}^{1/4}$ and ${\tau }_{\eta }=(\nu /ϵ{)}^{1/2}$. With $ϵ\sim u^3/\ell$, their ratios to the integral scales of turbulence are powers of the Reynolds number. They describe a mean scale balance; intermittent dissipative structures can have other local scales.

The dissipation potential $\Omega =\alpha \Vert Q{\Vert }^{n+1}/(n+1)-\mathrm{tr}(AQ)/\tau$, with the Frobenius norm, gives $\dot{A}+A/\tau =\alpha \Vert Q{\Vert }^{n-1}{Q}^T$. An integrating factor proves the displayed memory formula. For $n>0$, its nonlinear driving term is defined continuously as zero at $Q=0$. Mechanical dissipation is $\mathrm{tr}(Q\dot{A})=\alpha \Vert Q{\Vert }^{n+1}-\mathrm{tr}(AQ)/\tau$, which must be nonnegative on thermodynamically admissible histories.

The Lagrangian fluid displacement $\mathbf{\xi }$ compares neighbouring fluid trajectories. Linear velocity perturbations about a steady flow $\mathbf{U}$ obey $\delta \mathbf{u}=({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla )\mathbf{U}$. For a background scalar $f_0$, its material perturbation is $\Delta f=\delta f+\mathbf{\xi }\cdot \nabla f_0$. In an axisymmetric normal mode about cylindrical rotation, the radial and vertical velocities are $-i\omega {\xi }_r$ and $-i\omega {\xi }_z$. The displacement of a free surface is needed to impose the Lagrangian pressure perturbation condition.

// Target: astrophysics.bigb

For fixed complex frequency $\omega$, a spatial root is a complex wavenumber $k$ satisfying the dispersion relation. Analytic continuation of these branches from a causal temporal contour is essential in the Briggs-Bers criterion; their collision alone does not decide whether a spatial pinch point occurs.

With independent up moves of physical probability $p$ and risk-neutral probability $q$, the Radon-Nikodym derivative on the first $n$ periods is the displayed likelihood ratio, where $N_n$ counts up moves. Its conditional one-step multiplier has physical expectation one, so $L_k$ is a positive mean-one martingale. The terminal state-price density is $L_n/B_n$ for deterministic bank-account value $B_n$.

Enumerate all intervals with rational endpoints, using the countability of the rational numbers and listing pairs of enumeration indices by increasing sum. If every point $b$ of $B$ is an isolated point, some such interval contains $b$ and no other point of $B$. Assign to $b$ the least index of such an interval. Distinct points cannot have the same index, so this gives an injective function into $\mathbb{N}$ and proves that $B$ is a countable set. The argument uses isolation in the subspace topology, not closedness: the discrete subset $\{1/n:n\ge 1\}$ illustrates that an accumulation point outside $B$ is allowed.

A closed discrete subset $D$ of a topological space $X$ is a closed set whose subspace topology is the discrete topology. Every point of $D$ therefore has a neighbourhood meeting $D$ at that point alone. Closedness excludes all accumulation points outside $D$ as well.

In a plane domain, a closed discrete subset is exactly a set meeting every compact subset in finitely many points. This local finiteness is necessary for a nonzero holomorphic function to have the subset as its entire zero set. A merely isolated-point subset can have an accumulation point outside the subset but inside the domain, as $\{1/n:n\ge 1\}\subset \mathbb{C}$ does.

// Destination: analysis.bigb

A discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$, acting by Möbius transformations on the Riemann sphere or orientation-preserving isometries of hyperbolic space ${\mathbb{H}}^3$. A group may be Kleinian even when some elements fix points; freeness and absence of torsion are additional conditions.

A nonzero discrete additive subgroup of ${\mathbb{R}}^2$ has a basis of one or two real-linearly independent vectors. Choose a shortest nonzero vector $w_1$. If the subgroup is not contained in its line, reduce parallel components modulo $w_1$ and choose a vector $w_2$ of least positive perpendicular height. Such a height exists: heights tending to zero would give infinitely many reduced points in a bounded region, contrary to discreteness. Reducing any group vector modulo $w_2$ leaves height smaller than this minimum, hence height zero, and reducing along $w_1$ then gives an integer multiple of $w_1$.

A discrete additive subgroup of ${\mathbb{R}}^n$ is generated by real-linearly independent vectors. Indeed, isolation of zero gives uniform separation, so compact sets meet the subgroup finitely. Choose a real basis from the subgroup for its real span. Its integer span has finite index, since reduction into a compact fundamental parallelepiped gives finitely many coset representatives. The structure theorem for finitely generated abelian groups then gives a free integer basis, whose size is the dimension of the real span. Thus the subgroup is a Euclidean lattice in its span, although it need not have full rank in the ambient space.

The conditional union bound gives $\mathbb{E}{X}_i(k+1)\le \beta {\sum }_j{A}_{ij}\mathbb{E}{X}_j(k)$ in a discrete SIR epidemic on a graph. Iteration bounds the infection vector at step $k$ by $(\beta A{)}^{k}X(0)$. Each vertex is infected at most once, so summing the probabilities in time gives the displayed bound. Powers of the adjacency matrix of a graph count walks, which can overcount impossible reinfections without invalidating the upper bound. Initially removed individuals must be counted separately when bounding all eventual removals.

For $-\Delta u=f$ on a square with homogeneous Dirichlet boundary conditions, the five-point finite difference operator on a mesh $h=1/N$ is diagonalized by $\sin (\pi ki/N)\sin (\pi \ell j/N)$. Its eigenvalues are ${\lambda }_{k\ell }=4h^{-2}[{\sin }^2(\pi k/(2N))+{\sin }^2(\pi \ell /(2N))]$. A two-dimensional discrete sine transform, division by these nonzero eigenvalues and the inverse transform give the solution. Odd extension realizes each one-dimensional transform by a Fast Fourier transform of length $2N$, making the total cost $O(N^2\log N)$ and storage $O(N^2)$. These finite-difference eigenvalues should not be confused with the continuous spectral eigenvalues ${\pi }^2(k^2+{\ell }^2)$.

The sum of independent Bernoulli random variables with possibly unequal success probabilities $p_i$ has a Poisson binomial distribution. Its probability generating function is $G_K(z)={\prod }_i(1-p_i+p_{i}z)$: expand the product to read off the probability mass function. Linearity of expectation and additivity of variance for independent random variables give $\mathbb{E}K={\sum }_i{p}_i$ and $\mathrm{Var}(K)={\sum }_i{p}_i(1-p_i)$. Equal $p_i=p$ recover the binomial distribution.

For three trials with success probabilities $1/10,3/10,4/5$, expanding $(9/10+z/10)(7/10+3z/10)(1/5+4z/5)$ gives masses $0.126,0.572,0.278,0.024$ at $0,1,2,3$. Their expected value is $1.2$ and variance is $0.46$. Replacing all three success probabilities by their average $0.4$ preserves the expected value but gives binomial distribution variance $0.72$, so it does not preserve the law.