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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.

Given a smooth function $\kappa (s)$, setThen $\gamma$ is unit speed and has signed curvature $\kappa$. This gives the existence part of the fundamental theorem of plane curves.

The cohomology ring is the graded ring whose multiplication is the cup product. It records intersection information that the graded cohomology groups alone do not detect.

If distribution functions $F_n$ converge pointwise to a continuous distribution function $F$, then the convergence is uniform. Choose finitely many points whose successive $F$-increments and two tails are small, then use monotonicity to squeeze $F_n-F$ between its values at adjacent points.

The mean parameter corresponding to the natural parameter $\theta$ is

Differentiating the normalization identity gives

For a depressed cubic $X^3+aX+b=0$ over a field of characteristic zero, putChoose $U,V$ with $U^3=-b/2+\sqrt{D}$, $V^3=-b/2-\sqrt{D}$, and $UV=-a/3$. Then $x=U+V$ is a root, becauseFor $a\ne 0$, choose a nonzero $U$ and set $V=-a/(3U)$; this enforces the compatibility rather than choosing two arbitrary cube roots. If $a=0$, take a cube root of $-b$ directly. With a primitive cube root of unity $\omega$, the other roots are $\omega U+{\omega }^{2}V$ and ${\omega }^{2}U+\omega V$.

The Lagrange resolvent for a cubic explains why a quadratic equation appears inside the formula. Even when the cubic has three real roots, complex intermediate radicals may occur; the compatible pairing of cube roots is what makes the resulting expressions solve the original equation.

For elastic scattering of a plane wave by a periodic crystal, a diffraction contribution can occur when the momentum transfer is a reciprocal lattice vector:Squaring ${\mathbf{k}}^{\prime }=\mathbf{k}-\mathbf{q}$ giveswhere $\theta$ is the angle between incident and outgoing wavevectors. This is the Ewald sphere construction. For planes with spacing $d$ and first reciprocal spacing $2\pi /d$, it becomes Bragg's law $2d\sin (\theta /2)=\lambda$.

The geometrical condition alone does not ensure an observed reflection. The structure factor must also be nonzero: different atoms in a motif can cancel the scattering amplitude, producing a systematic absence. The diamond-basis diffraction extinction is a concrete example.

For a powder or a crystal rotated to reach the required orientation, the smallest possible nonzero diffraction angle is therefore determined by the shortest allowed reciprocal lattice vector with nonzero structure factor and $|\mathbf{q}|\le 2k$. For a crystal at one fixed orientation, the incident direction additionally determines which reciprocal vectors satisfy the displayed elastic scattering condition.

For a finite set $S$ of crystal sites, the lattice structure factor at momentum transfer $Q$ isIt multiplies the single-atom scattering amplitude.

A three-dimensional Bravais lattice iswhere the primitive vectors $a_1,a_2,a_3$ are linearly independent.