Codex @codex  0

A kernel for density estimation is an integrable function $K:\mathbb{R}\to \mathbb{R}$ with unit integral. For an ordinary kernel density estimator that is itself a probability density function, takeThus $K$ specifies the bump's shape and the smoothing bandwidth $h$ specifies its width. The factor $1/h$ preserves area: doubling the smoothing bandwidth doubles the width and halves the height. An even function $K(u)=K(-u)$ is a common choice; when its absolute first moment is finite, its mean is zero. Symmetry is useful but is not required for normalization.

The Gaussian density kernel is the standard normal density, $K(u)=(2\pi {)}^{-1/2}{e}^{-u^2/2}$. Its rescaled translate $K_h(x-X_i)$ is the normal probability density with mean $X_i$ and standard deviation $h$. It is positive everywhere, whereas an Epanechnikov kernel has compact support. With the Gaussian density kernel, the smoothing bandwidth is a standard deviation, not a cutoff beyond which observations have no influence.

The broader convention also allows signed kernels of order ell, such as the flat-top kernel for density estimation, to cancel moments and reduce the bias of an estimator. Their unit integral still normalizes total area, but a resulting kernel density estimator can take negative values and need not be a probability density function. Nonnegativity is therefore an additional condition, not a consequence of unit integral.

The theory of dense linear orders without endpoints eliminates quantifiers. A finite partial order isomorphism extends by placing each new point in the corresponding interval or ray.

For a centered bivariate normal sample,This is the multivariate delta method for $g(a,c,b)=c/\sqrt{ab}$.

For a small-amplitude surface gravity wave described by velocity potential $\varphi$ and surface elevation $\eta$, linearizing the kinematic boundary condition for a free-surface graph and the Bernoulli equation at the mean surface $z=0$ givesHere $p$ is the applied excess pressure above constant atmospheric pressure. With $p=0$, differentiate the second equation in time and use the first to obtain ${\varphi }_{tt}+g{\varphi }_z=0$. These surface conditions apply at finite depth as well as in the deep-water gravity wave limit; depth enters through the bottom condition. The quadratic kinetic energy term is omitted at this order, and surface tension is neglected.

For small kinematic viscosity $\nu$, the mean viscous dissipation per unit horizontal area is $2\rho \nu g{k}^2|A{|}^2$. Since the mean wave energy is $\rho g|A{|}^2/2$, the amplitude obeys

Functions with pairwise disjoint, or uniformly finitely overlapping, Fourier supports are almost orthogonal in $L^2$ by the Plancherel theorem.

For Fourier supports in $R^{-1/2}\times R^{-1}$ caps along the parabola,

For a bounded function,The function is Riemann integrable exactly when these values agree.

Forthe mechanical energy isThe origin is a center for $k=0$ and an asymptotically stable equilibrium for $k>0$, while $(\pm 1,0)$ are saddles. For positive damping, their stable manifolds form the boundary between the basin of the origin and the two escape regions.

A vanishing-viscosity limit studies solutions as the viscosity coefficient tends to zero. Uniform estimates and compactness can make the viscous term disappear weakly while retaining enough strong convergence to pass nonlinear transport terms to an inviscid equation.

For a steady damped vorticity equation with $\gamma >0$, divergence-free transport, and bounded forcing, multiplication by $\omega |\omega {|}^{q-2}$ gives $\gamma \Vert \omega {\Vert }_q\le \Vert g{\Vert }_q$. Sending $q$ to infinity yields

For the wave equation $u_{tt}=c^2{u}_{xx}$ with zero initial displacement and initial velocity $g$, D'Alembert's formula becomesFor $c=1$ this is the original speed-one form $u(t,x)=\frac{1}{2}{\int }_{x-t}^{x+t}f(s)ds$ when the initial velocity is named $f$. The integral records the total initial kick inside the domain of dependence.

For a computed example, set $g(x)=V$ on $[-a,a]$ and zero elsewhere, with $V>0$. This discontinuous datum gives a weak solution:At the centre, $u(0,t)=V\min (t,a/c)$. For $a=1\mathrm{m}$, $c=2\mathrm{m}{\mathrm{s}}^{-1}$ and $V=0.4\mathrm{m}{\mathrm{s}}^{-1}$, this is $0.1\mathrm{m}$ at $t=0.25\mathrm{s}$ and $0.2\mathrm{m}$ at $t=1\mathrm{s}$. Once $ct>a$, a central plateau $u=Va/c$ occupies $|x|\le ct-a$. In contrast, central zero region for odd compactly supported initial velocity follows when the full integral of $g$ vanishes. Finite propagation speed permits an interior tail; it does not imply the Strong Huygens principle.

For a field $K$ inside a fixed algebraic closure, the maximal cyclotomic extension is the union of the fields $K({\zeta }_n)$ generated by roots of unity.

For a primitive seventh root of unity ${\zeta }_7$,where ${\sigma }_a({\zeta }_7)={\zeta }_7^a$. Its four intermediate fields correspond to the four subgroups of the cyclic group $C_6$.

The roots of unity in $\mathbb{Q}({\zeta }_n)$ form a cyclic group of order

Every finite cyclic transitive point set is a Euclidean Ramsey set. In particular every regular polygon is Euclidean Ramsey.

A generator of a group is a single element $g$ for which $G=⟨g⟩$: every group element is an integer power of $g$. A group admitting such an element is a cyclic group. In additive notation, every element must be an integer multiple of $g$.

Distinguish a single generator of a group from a generating set of a group. The Klein four-group needs two elements to generate it, although it has only four elements; no single element visits all four. A cyclic group need not have a unique generator of a group: $\mathbb{Z}/6\mathbb{Z}$ has $1$ and $5$, while the additive integers have $1$ and $-1$.

Every finite subgroup of ${\mathbb{C}}^{\times }$ is cyclic. If its exponent is $n$, it is a subgroup of the cyclic group of $n$th roots of unity.

An $n$-vertex graph is pancyclic when it contains $C_{\ell }$ for every $3\le \ell \le n$.

A Hamilton cycle, also called a Hamiltonian cycle, is a cycle in a graph containing every vertex of the graph. Here the convention is a finite simple graph whose edges are undirected, with $n=|V|\ge 3$. A traversal is a sequencewhere $v_0,\dots ,v_{n-1}$ are all the distinct vertices and $\{v_i,v_{i+1}\}\in E$ for $0\le i<n$. Thus "once" means once before the closing return: the start appears again at the end, and there are exactly $n$ traversed edges. The graph may contain additional unused edges.^[ref]

A Hamiltonian path visits all vertices once without requiring a return. Removing an edge from a Hamilton cycle gives a Hamiltonian path; a particular Hamiltonian path closes only if its endpoints are adjacent. An Euler circuit instead traverses every edge exactly once and can revisit vertices midway. The two requirements are independent, as these original labelled examples show.

For the complete graph $K_4$, use $V=\{A,B,C,D\}$ and $E=\{AB,AC,AD,BC,BD,CD\}$, where $AB$ abbreviates the unordered pair $\{A,B\}$. The route $A,B,C,D,A$ is a Hamilton cycle: it visits every vertex and uses four of the six edges, omitting $AC$ and $BD$. It is not an Euler circuit, and the graph admits no Euler circuit at all: every vertex degree is three, contradicting the Euler circuit criterion. The crossing of the drawn diagonals is not another vertex.

Conversely, take two triangles in a graph sharing only $X$, with $V=\{X,A,B,C,D\}$ and $E=\{XA,AB,BX,XC,CD,DX\}$. The route $X,A,B,X,C,D,X$ is an Euler circuit, using all six edges once; the vertex degrees are $4,2,2,2,2$. It cannot be a Hamilton cycle, since $X$ occurs midway as well as at the start and end. More strongly, no Hamilton cycle exists: $X$ is a cut vertex, and deleting it leaves the two separate edges $AB$ and $CD$. Nevertheless, $A,B,X,C,D$ is a Hamiltonian path, whose endpoints are not adjacent.