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Write $a=D/2$ and wavenumber $k=2\pi /\lambda$. The pupil Fourier transform, normalized by pupil area, gives the Fraunhofer diffraction wave amplitude. In pupil polar coordinates,The angular average is $J_0$ because $J_0(z)=(2\pi {)}^{-1}{\int }_0^{2\pi }{e}^{-iz\cos \varphi }d\varphi$. The Bessel function of the first kind identity $d\{z{J}_1(z)\}/dz=z{J}_0(z)$ then reduces the radial integral to $A(u)=2J_1(u)/u$. Taking the squared magnitude yields the Airy disk light intensity profile.^[ref]

The apparent division by zero is removable. The power seriesgives $A(u)=1-u^2/8+u^4/192+O(u^6)$ and $P(u)=1-u^2/4+5u^4/192+O(u^6)$. At the centre, every pupil contribution has the same wave phase, so their normalized sum is one. The first dark ring instead corresponds to $J_1(u)=0$ with $u>0$, rather than to the removable zero of $J_1$ at the origin.

An affine map need not be injective. For $G(x,y)=(x+y,0)$, the direction $(1,-1)$ belongs to the kernel of its linear part of an affine map. Consequently the whole line $y=-x$ maps to $(0,0)$, and every parallel line $y=-x+c$ maps to $(c,0)$. These are collapsed lines, not images with a surviving direction. Assertions that an affine map preserves distinct parallel lines require the appropriate noncollapse or injectivity hypothesis.

For coefficients with sum one, the defining identity follows from linearity:The translation appears once on both sides precisely because ${\sum }_i{\lambda }_i=1$. In particular, midpoints map to midpoints, and a parameterized line segment $(1-t)P+tQ$, $0\le t\le 1$, maps to $(1-t)T(P)+tT(Q)$. A whole line has the same formula with $t\in \mathbb{R}$; its image is a line if $M(Q-P)\ne 0$, and a single point otherwise.

Keep the basis fixed and replace the origin by a point with old coordinates $o$. Each point's coordinates change from $p_i$ to $p_i-o$. For coefficients with sum one, the weighted coordinates becomeAdding the new origin $o$ gives the original point. This proves origin independence rather than asserting that the coordinate tuples themselves stay unchanged.

The requirement cannot be omitted. In the example $A=(0,0)$, $B=(2,0)$, blindly treating $A+B$ as a point gives $(2,0)$ in the old coordinates. If the new origin is $o=(1,1)$, the two coordinates become $(-1,-1)$ and $(1,-1)$; adding them and restoring $o$ instead gives $(1,-1)$. The coefficients sum to two, so this procedure has no origin-independent interpretation as a weighted point.

An alpha-adrenergic receptor belongs to the alpha branch of adrenergic receptors, divided into alpha-1 adrenergic receptors and alpha-2 adrenergic receptors. The branches differ in their common G-protein coupling and intracellular signals. Grouping them as alpha receptors identifies related pharmacology, but does not mean that all produce the same response or activate the same messenger pathway. ^[ref]

A beta-adrenergic receptor is an adrenergic receptor subtype family commonly coupling to stimulatory G-protein signalling and increased cyclic AMP. Cellular responses include changes in cardiac activity or fuel mobilization, depending on subtype and tissue. Ligands can stabilize different receptor conformations, so binding affinity alone does not determine efficacy or signalling preference. ^[ref]

Choose a strict total order $v_1<\cdots <v_n$ of the vertices of a finite simple graph. Direct every edge $v_i{v}_j$ from $v_i$ to $v_j$ when $i<j$. Along every directed path, the index strictly increases. Returning to the starting vertex would require an impossible strict increase back to the initial index, so this is an acyclic orientation. The displayed $D_1$ comes from $A<B<C$.

The chosen strict total order is a topological order of the resulting directed acyclic graph. Conversely, every acyclic orientation of a finite simple graph has a topological order, and orienting its edges from earlier to later in that order recovers it. Different topological orders can induce the same acyclic orientation, since vertices with no edge between them need not impose a choice of arrow.^[ref]

An order respects reachability but does not create it: $u<v$ in a chosen topological order does not imply a directed path from $u$ to $v$. For example, orient the two edges of the graph path $A,B,C$ using $A<C<B$. The arrows are $A\to B$ and $C\to B$; there is no directed path from $A$ to $C$, despite $A<C$. In $D_1$, however, $A$ reaches $C$ both directly and via $B$. Following outgoing directed edges with graph search decides this directional reachability.^[ref]

For a square integer matrix $A$ with nonzero determinant, the image $A{\mathbb{Z}}^n$ is a subgroup of the integer lattice. Its index of a subgroup isIn particular, $A=\mathrm{diag}(2,1)$ gives all points with even first coordinate. The quotient has two cosets, represented by $(0,0)$ and $(1,0)$, and the fundamental cell has area two. A determinant of two therefore counts how much integer information is missing from the image.

The integrality hypothesis is essential. The rational matrix $R=\mathrm{diag}(2,1/2)$ has determinant one but sends $(0,1)$ to $(0,1/2)$. Its generated Euclidean lattice has the same covolume as ${\mathbb{Z}}^2$ without being the same integer lattice. The unimodular matrix criterion combines an arithmetic condition with a geometric one.

A short proof of the index formula uses Smith normal form. Write $PAQ=\mathrm{diag}(d_1,\dots ,d_n)$, with $P,Q$ unimodular matrices and positive integers $d_i$. They preserve the relevant integer structure, so the quotient has the same size as ${\prod }_i\mathbb{Z}/d_i\mathbb{Z}$, namely ${\prod }_i{d}_i=|\det A|$. No assumption that the matrix is orthogonal or its basis vectors are short is needed.

Let $B_1,B_2$ be real square invertible matrices whose columns are lattice bases. They generate the same Euclidean lattice precisely when^[ref]

If $B_2=B_{1}U$, multiplication by $U$ merely permutes the integer coordinate vectors, so the two generated sets agree. Conversely, equality of the generated sets means every column of either basis is an integer linear combination of the other. Hence $B_2=B_{1}U$ and $B_1=B_{2}V$ for integer matrices $U,V$. Substitution followed by multiplication by $B_1^{-1}$ gives $UV=I$, so the integer inverse criterion for a unimodular matrix applies.

Equal real linear spans are a weaker condition: any two real square invertible matrices span the same ambient vector space. Even equal covolumes do not force the same Euclidean lattice. The requirement that the change matrix and its inverse matrix have integer entries provides the missing arithmetic information.

For a square integer matrix $U$ with nonzero determinant, the inverse matrix has integer entries exactly when $\det U=\pm 1$. The adjugate matrix formula proves one direction:Every entry of $\mathrm{adj}(U)$ is an integer cofactor, so division by $1$ or $-1$ leaves integer entries. Conversely, if both $U$ and $U^{-1}$ are integer matrices, their determinants are integers with product one. The only possibilities are that both are one or both are minus one. This explains why a general integer matrix with nonzero determinant need not have an integer inverse matrix.

Multiple edges are distinct edges with the same pair of endpoint vertices in an undirected graph. A multigraph permits them; a simple graph does not. In an orientation of a graph of a multigraph, each such edge receives its own direction, so two of them can supply opposite directed edges between the same vertices.

For a specified optical transition, the on-resonance saturation parameter is the light intensity divided by the saturation intensity. The independent-beam Doppler cooling force model assumes weak total excitation and defines $s$ separately for each beam.

For a convergent series with sum $S$, the error from keeping only the first $N$ terms is the remaining sum $R_N={\sum }_{n>N}{a}_n$. This truncation error exists even with exact arithmetic. The zeta series truncation error gives explicit bounds for positive reciprocal-power terms.

A sequence $(x_n)$ in a topological space converges to $x$ if every open neighbourhood of $x$ contains all sufficiently late terms:For a metric space, this is the usual limit of a sequence condition expressed through open balls. For a Hausdorff space, distinct proposed limits have disjoint open neighbourhoods, so a sequence cannot eventually stay in both and the limit is unique. The sequence convergence in the cofinite topology demonstrates what can change when that separation property is absent.^[ref]

The category of topological manifolds has topological manifolds without boundary as its objects of a category and all continuous maps between them as its categorical morphisms. Allowing every dimension of a manifold lets maps connect spaces of different dimensions. Identity functions are continuous, and function composition preserves continuity, so the category laws are inherited from functions. Its isomorphisms are exactly homeomorphisms.

In coordinate charts $(U,\phi )$ on $M$ and $(V,\psi )$ on $N$, a continuous map $f:M\to N$ is represented locally by $\psi \circ f\circ {\phi }^{-1}$ on $\phi (U\cap f^{-1}(V))$. Changing coordinate charts changes this expression through topological transition maps, while $f$ itself stays fixed. This viewpoint separates the underlying continuous map from the coordinates used to describe it; restricting to smooth manifolds and smooth maps gives the corresponding smooth category.

For $n\ge 1$, a topological $n$-manifold with boundary is a Hausdorff second-countable space locally homeomorphic to relatively open sets in the half-space ${\mathbb{H}}^n$. The openness here is in the subspace topology of ${\mathbb{H}}^n$, not necessarily in ${\mathbb{R}}^n$. Its intrinsic boundary $\partial M$ consists of points sent to $x_n=0$; the remaining points have ordinary coordinate charts with images open in Euclidean space. The intrinsic boundary is independent of the chosen chart.^[ref]

The closed interval $[0,1]$ illustrates the distinction: its interior points have open interval charts, while $0$ and $1$ have charts modeled on $[0,\epsilon )\subseteq {\mathbb{H}}^1$, using $x$ and $1-x$. Its intrinsic boundary is $\{0,1\}$. The circle has empty intrinsic boundary even though its boundary of a set as a subset of the Euclidean plane is the whole circle. A topological manifold without boundary is therefore a special case with $\partial M=\varnothing$, and a smooth atlas on the half-space charts gives a manifold with boundary in the smooth sense.

Take the circle $S^1=\{(x,y)\in {\mathbb{R}}^2:x^2+y^2=1\}$ with its subspace topology. Write $W=(-1,0)$ for its west point and $E=(1,0)$ for its east point. Two coordinate charts obtained by stereographic projection areBoth take values in the entire real line. Geometrically, join $W$ to $(x,y)$ and intersect the resulting line with $x=0$: the height is $t$. Projecting from $E$ instead gives height $u$. Each projection omits its own starting point, but the other coordinate chart includes that point. Hence $U\cup V=S^1$.^[ref]

To verify the homeomorphisms, solve the displayed equations for $(x,y)$. The inverse functions areEach formula lies on the circle, avoids its excluded point, and is continuous for every real argument. Substitution into the projection formulas returns $t$ or $u$; substitution in the other direction returns $(x,y)$. Thus these are continuous bijections with continuous inverse functions, as required for coordinate charts. Since the Euclidean plane is Hausdorff and second countable, the subset $S^1$ with its subspace topology is too. This checks all three conditions for a topological manifold, rather than just drawing a smooth-looking curve.

On the overlap $U\cap V$, both $x=\pm 1$ are excluded, so $y\ne 0$. Using $y^2=1-x^2$, the topological transition map isFor example, $(x,y)=(3/5,4/5)$ has $t=1/2$ and $u=2$. The two connected components of the overlap, the upper and lower arcs, correspond to the positive and negative parts of the punctured real line. The values $t=0$ and $u=0$ represent $E$ and $W$, respectively, and belong to only one chart each. The reciprocal coordinate change and its inverse function are smooth functions away from zero, so this topological atlas also supplies a smooth atlas.

Why can neither chart be enlarged to a single coordinate chart on the whole circle? The circle is a closed set and a bounded set in the Euclidean plane, hence compact by the Heine-Borel theorem. A global coordinate chart would give a homeomorphism to a nonempty open set $O\subseteq \mathbb{R}$. Its image would be compact and therefore closed in the real line. By connectedness of the real line, this forces $O=\mathbb{R}$, which is not compact. This is the one-dimensional instance of no compact manifold has a single Euclidean chart. Locally the circle is indistinguishable from the real line; globally compactness obstructs a single chart.