For a first-order ordinary differential equation rational in the dependent variable, with coefficients meromorphic on the chosen independent-variable surface, analytic continuation along an arc ending outside the fixed exceptional set has a definite limit in the Riemann sphere. Its movable singularities are poles or algebraic branch points. To see the limiting assertion, the cluster set of a continuous lifted arc is connected and compact. Any cluster value where the vector field is holomorphic permits uniform local continuation by the Picard-Lindelöf theorem. Otherwise the cluster set lies in finitely many coefficient poles, so it is a singleton. At a pole of the vector field, the inverse equation is holomorphic; its solution has a finite-order critical point, giving a convergent root expansion. The reciprocal dependent variable handles infinity. Points where both reduced numerator and denominator vanish belong to the fixed exceptional set and are excluded from this argument.

The Painlevé property excludes movable critical singularities, such as branch points, of the analytically continued general solution of a complex differential equation. A common stronger formulation requires every movable singularity to be a pole, so local solutions are meromorphic there. Fixed singularities determined by the coefficients are distinguished from movable ones whose locations depend on initial data. Riccati equations with meromorphic coefficients have only movable poles at ordinary points: linearization expresses their solutions as logarithmic derivatives of linear-system solutions. More generally, set $w=f/g$ and solve the holomorphic linear system $f^{\prime }=bf+cg$, $g^{\prime }=-af$. Its quotient satisfies $w^{\prime }=a{w}^2+bw+c$ directly. Initial data $g(z_0)=1$, $f(z_0)=w(z_0)$ recover every local solution, and holomorphic $f,g$ make the quotient a meromorphic function, even at zeros of $a$. At a zero of $g$ with $a\ne 0$, uniqueness prevents $f=0$ there, so $g^{\prime }=-af\ne 0$ and the pole is simple. This proves the absence of movable branching without assuming that the quadratic coefficient never vanishes.

Two complex lattices are homothetic when multiplication by one nonzero complex number takes one onto the other. Changes between oriented integral bases have determinant one, so normalized period ratios classify these classes by $S{L}_2(\mathbb{Z})\backslash \mathbb{H}$. The lattice Eisenstein sums scale as $G_k(a\Lambda )=a^{-k}{G}_k(\Lambda )$.

For the complex-linear Hodge star operator and complex orientation on a Hermitian manifold of complex dimension $n$, every $(n,0)$-form has the displayed eigenvalue. In an adapted orthonormal real coframe, each factor $d{x}_j+id{y}_j$ has two-dimensional star eigenvalue $-i$. Regrouping the $n$ factors introduces $(-1{)}^{n(n-1)/2}$, giving $(-1{)}^{n(n+1)/2}{i}^n$. This is a pointwise metric identity; the metric need not be Kähler.

Use the complex-linear Hodge star operator on the direct sum of all form degrees. Its square on degree $r$ is $(-1{)}^{r(n-r)}$. In positive odd dimension this is always one; in positive even dimension its sign is positive on even-degree forms and negative on odd-degree forms. Complementary basis wedges provide both roots of each sign, so the displayed eigenvalues all occur. On a single degree the Hodge star operator is an endomorphism only in middle degree; on middle degree $m$ in dimension $2m$ its possible eigenvalues are $\pm 1$ for even $m$ and $\pm i$ for odd $m$.

For real $\mu$ and $\alpha >0$, this complex amplitude equation has a destabilizing cubic term and a saturating quintic term. Nonzero uniform intensities solve $\mu +\alpha q-q^2=0$. The upper branch is radially stable and the lower branch radially unstable; constant phase is neutral because multiplication by $e^{i{\varphi }_0}$ preserves the equation. Stable conduction and the stable upper pattern coexist for $-{\alpha }^2/4<\mu <0$. At the fold, a zero radial eigenvalue conceals one-sided nonlinear instability, so strict stability cannot be inferred from eigenvalues alone.

For $A_t=A+(1+ib)A_{xx}-(1+ic)|A{|}^{2}A$, the uniform oscillation $A=e^{-ict}$ has phase diffusion coefficient $D=1+bc$. Its amplitude-phase dispersion relation is ${\lambda }^2+(2+2k^2)\lambda +2D{k}^2+(1+b^2)k^4=0$. Thus $D<0$ gives a long-wavelength modulational instability, whereas $D>0$ gives linear stability of all nonzero Fourier modes. On a fixed periodic domain, instability requires an allowed wavenumber with $k^2<-2D/(1+b^2)$.

The orientation of a contour specifies the direction in which its parametrization is traversed. Reversing that direction negates its contour integral. For a simple closed curve in the complex plane, positive orientation of a contour is the direction keeping its bounded interior on the left.

A piecewise smooth curve has a continuous function as a parametrization on a compact interval, with a finite partition into subintervals on each of which the parametrization is continuously differentiable. One-sided derivatives may differ at joining points, so polygonal contours are allowed. A contour integral is the sum of the integrals over those smooth pieces.

Choose compatible primitive roots of unity ${\zeta }_{p^{n+1}}$. The partial sums $A_N={\sum }_{n=0}^N{p}^{2n}{\zeta }_{p^{n+1}}$ form a Cauchy sequence. Each $A_N$ generates ${\mathbb{Q}}_p({\zeta }_{p^{N+1}})$: for a nonidentity cyclotomic automorphism with $r=v_p(a-1)$, the first changed term has valuation $2r+1/(p-1)$, strictly below every later changed term. Thus the conjugate distance is at least $p^{-2N-1/(p-1)}$. The tail bound is $p^{-2N-2}$, strictly smaller. If the limit were algebraic, Krasner's lemma would place every one of these unbounded-degree cyclotomic fields in one finite extension, a contradiction.

A completely positive map with ${\sum }_{\alpha }{K}_{\alpha }^{†}{K}_{\alpha }\le I$ is a trace-nonincreasing quantum operation, representing an outcome branch with its probability retained in the trace. A deterministic quantum channel has equality and is trace preserving. Dividing a branch output by its trace gives a conditional state but generally produces a nonlinear update; that normalization is not part of the linear quantum operation.

Bounded completeness requires the defining implication for a complete statistic only for bounded measurable functions. Completeness implies it, but the converse need not hold. If a sufficient statistic $T$ gives a parameter-independent event probability $c$ and the conditional probability $S(T)$ of that event is nonconstant, then $S-c$ is a bounded zero-mean obstruction to bounded completeness.

A metric space is complete if and only if every decreasing sequence of nonempty closed sets whose diameters tend to zero has nonempty intersection. In that case the intersection contains exactly one point.

For two Cauchy sequences, $|d(x_n,y_n)-d(x_m,y_m)|\le d(x_n,x_m)+d(y_n,y_m)$, so the displayed limit exists. It defines a pseudometric on the Cauchy sequences, and its metric quotient identifies sequences whose mutual distance tends to zero. Constant sequences embed the original metric space isometrically. Their image is dense, because the class of $(x_n)$ is approached by the constant classes of $x_n$. The quotient is complete: from a Cauchy sequence of its points select a subsequence whose successive distances are below $2^{-j}$ and choose original-space points within $2^{-j}$ of these subsequence points. Those original-space points form a Cauchy sequence and its class is the limit of the subsequence, hence of the whole sequence. The original embedding is onto exactly when the original space is complete.

In a one-period complete market with a strictly positive numéraire, at most one positive state-price density prices every traded asset. Suppose $\rho ,{\rho }^{\prime }$ do so. For each event $A$, replicate the contingent claim $S_1^{(0)}{1}_A$. Linearity of the pricing equations givesBoth weighted densities are integrable because they price the numéraire. An integrable random variable whose integral over every event is zero vanishes almost surely: take the events where it is positive and negative. Therefore $(\rho -{\rho }^{\prime })S_1^{(0)}=0$, and strict positivity of the numéraire gives $\rho ={\rho }^{\prime }$. This argument avoids assuming that the unweighted densities themselves are integrable.

In an arbitrage-free one-period market with finitely many assets and an equivalent martingale measure, market completeness is equivalent to uniqueness of the dominated martingale measure. Completeness prices every indicator function, forcing two measures to agree on every event. Conversely, if the attainable payoff span has dimension $k$ but is not the whole payoff space, choose $k+1$ disjoint positive-probability events. Their moment vectors against a basis of attainable payoffs are linearly dependent. A nonzero bounded linear combination $h$ of their indicators consequently has zero pricing moments. Multiplying an existing equivalent density by $1+\epsilon h$, for sufficiently small positive and negative $\epsilon$, constructs distinct equivalent martingale measures. This proves the converse without assuming that the original space was finite. On a general probability space, completeness here means replication of all bounded claims, or equivalently all square-integrable claims when asset payoffs are square-integrable.

A star graph is the complete bipartite graph with one centre and $m$ leaves. Any pair of its edges meet at the centre.

Fix a $(p+1)$-set $B\subseteq [4p]$. Take every $2p$-set $B\cup X$ with $X$ a $(p-1)$-subset of the remaining $3p-1$ points, together with the complements of all these sets. Within either half, intersections have size at least $p+1$. Across the halves they have size $|X\setminus Y|\le p-1$. Thus no intersection has size $p$, and the two disjoint halves give the displayed family size.

A universal set $U$ fixes the objects under consideration in a particular discussion; each set under consideration is a subset of $U$. For example, $U=\{1,2,3,4,5,6\}$ makes the complement of $\{1,2,3\}$ equal to $\{4,5,6\}$. This use of “universal” is local to the discussion: it does not assert the existence of a set containing every set. Stating $U$ makes exclusion rules precise.

Let eventual outcome $I=j$ have probability ${\pi }_j$, with $\log T\mid I=j\sim N({\mu }_j,{\sigma }_j^2)$. An observed outcome at time $t$ contributes ${\pi }_j{f}_j(t)$ to the likelihood, while right censoring at $c$ contributes ${\sum }_j{\pi }_j{S}_j(c)$. The expectation-maximization algorithm imputes both the censored outcome and its latent log event time. Its E-step uses the displayed conditional weights and the first two moments of a truncated normal distribution above $\log c$. With $z=(\log c-{\mu }_j)/{\sigma }_j$ and inverse Mills ratio $\psi (z)$, these moments are $m={\mu }_j+{\sigma }_j\psi (z)$ and $s=m^2+{\sigma }_j^2[1-\psi (z)(\psi (z)-z)]$. Observed outcomes use deterministic class weights and moments $\log t,(\log t{)}^2$. The M-step updates each class probability to its average weight, its mean to the weighted first moment, and its variance to the weighted second moment minus the squared new mean. Outcome-specific cumulative probabilities are cumulative incidence functions, not the conditional distributions $F(t\mid I=j)$ themselves.