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Can two readout values locate a hit between wires? Only with a specified response model. Place two readout wires at $0$ and $p$, put $u=x/p\in [0,1]$, and stipulate calibrated, noiseless electric chargesThese are assumptions for a toy sensor response, not the natural response law of adjacent anode wires. Under these assumptions the centroidrecovers position exactly. The apparent gain in precision comes from adding calibrated analogue information. Taking the mean of two fired wire addresses without their electric charges would instead give the fixed midpoint $p/2$.

For a chosen noise model, replace the two electric charges by $q_L+n_L$ and $q_R+n_R$, with independent random variables $n_L,n_R$ of zero expected value and common standard deviation $s$. Assume $s/Q\ll 1$ and no threshold censoring. First-order expansion of the ratio givesFor stipulated $p=2\mathrm{mm}$ and $s/Q=0.02$, this approximation gives $0.0283\mathrm{mm}$ at the midpoint and $0.0400\mathrm{mm}$ at either endpoint. The endpoint values describe uncensored signed channel readouts, including small negative values after baseline subtraction. They do not assume that both channels would pass a positive threshold. A ratio with a fluctuating denominator also need not have exactly zero bias; this is a local small-noise approximation.

Three original counterexamples show why calibration and selection matter:

Likewise, two equal hits at $u=1/4$ and $u=3/4$ have exactly the same total two-channel response as one hit at $u=1/2$ with twice the electric charge. A centroid can hide multiplicity even when each charge measurement is exact.

Diffusion enabled the lecture's Orsay interpolation, without establishing this model's linear response.^[ref]

Every point in the convex hull of a subset of ${\mathbb{R}}^d$ is a convex combination of at most $d+1$ points of that subset. Thus in the plane three ingredients always suffice, however many samples were supplied. The representation of $p=(2,1)$ in convex hull illustrates this bound. Two ingredients need not suffice: the centroid of a triangle with only its three vertices as input is on none of their three pairwise line segments.

A short proof outline removes redundant ingredients. With more than $d+1$ points, the augmented vectors $(s_i,1)$ are linearly dependent. A dependence lets the weights vary while preserving both their sum and the represented point. Move along it until one positive weight first becomes zero; all weights stay nonnegative. Repeating leaves at most $d+1$ positive weights.

In his 1911 paper, Constantin Carathéodory used convex geometry to investigate the possible Fourier coefficients of positive harmonic functions. Its introduction interprets convex mixtures through the center of mass, connecting finite-dimensional geometry to constraints on a function's coefficients.^[ref]

The centre of buoyancy is the point through which the resultant hydrostatic buoyant force acts. In a uniform fluid under uniform gravitational acceleration, it is the volume centroid of the displaced fluid. For a floating body its position generally changes when the body tilts, because the displaced volume changes shape.

For a nonempty finite collection of Euclidean vectors, its sample mean $\bar{x}=n^{-1}{\sum }_i{x}_i$ uniquely minimizes the sum of squared Euclidean distances to a proposed centroid $z$. Expand each squared Euclidean norm around $\bar{x}$ and use ${\sum }_i(x_i-\bar{x})=0$ to cancel the cross term. The displayed identity leaves a nonnegative term $n\Vert z-\bar{x}{\Vert }^2$, equal to zero exactly at $z=\bar{x}$.

On the probability simplex, ${\sum }_i{x}_i=1$. The Cauchy-Schwarz inequality gives $1\le \sqrt{n}\Vert x\Vert$, with equality only when all coordinates are equal. Thus the displayed centroid is the unique closest point to the origin. It is also equidistant from all coordinate vertices, since $\Vert x-e_i{\Vert }^2=\Vert x{\Vert }^2-2x_i+1$.

A convex combination is a finite weighted average of points in a real vector space. For $n\ge 1$, it has the formThe nonnegative coefficients forbid subtracting a component, and their unit sum keeps the result on the same affine scale as the inputs. With two points it parametrizes their line segment; equal coefficients give their centroid. A convex set contains every finite convex combination of its points. Conversely, the two-point case already gives the defining property of a convex set.^[ref]

The distinction from a linear combination is concrete. Combining $(0,0)$ and $(2,0)$ with coefficients $-1$ and $2$ gives $(4,0)$, outside their line segment, despite the unit coefficient sum. Using two coefficients both equal to one gives $(2,0)$ here, but their sum is two, so it is not a convex combination with those coefficients. A numerical output may happen to coincide with a point admitting a different convex representation; the definition applies to the specified coefficients.

An earthquake centroid is an effective average location of seismic moment release. It differs from the hypocenter, where rupture starts, and can depend on the source representation and observations. An associated centroid time similarly averages release rather than marking its first instant.

For a finite dataset, alternate nearest-centre assignment and replacement of each centre by the centroid of its assigned observations. Keep the current assignment whenever an observation is tied between nearest centres. A changed assignment then strictly decreases the within-cluster sum of squared Euclidean distances; updating centres cannot increase it by the centroid sum-of-squares identity.

After each centre update, the centres are determined by the current partition of a set of observation indices. There are only finitely many such partitions, so a strictly improving sequence cannot revisit one and must stop. A convenient exact stopping test is to check after each update whether every observation is still assigned to a nearest centre. This proof is for algorithm termination; it does not guarantee a global minimum.

Empty clusters need an explicit convention. To retain exactly $K\le n$ nonempty clusters, transfer an observation from a cluster with at least two members to each empty cluster, making it a singleton centred at that observation. This cannot increase the objective: the transferred observation now has squared distance zero, and no other contribution changes before centre updating. A deterministic choice is an eligible observation farthest from its currently assigned centre, with ties resolved by index. After the first round, an empty cluster can arise only during a strictly improving reassignment, so these repairs preserve the finite-termination argument.

For a planar region $R$ of finite positive area $A$, its first moments of area about the coordinate axes are the displayed double integrals, provided they exist. Its centroid is $(\overline{x},\overline{y})=(M_y/A,M_x/A)$. This follows by applying the definition $\overline{\mathbf{r}}=A^{-1}{\iint }_R\mathbf{r}dA$ to each component. The subscript indicates the axis about which the moment is taken: the first moment about the $x$-axis weights by $y$, not $x$.

A left group action of a group $G$ on a set $X$ is a function $(g,x)\mapsto g\cdot x$ satisfyingwhere $e$ is the identity element of $G$. The second rule means that $h$ moves the point first and $g$ moves it next. These rules let group multiplication describe consistent rearrangements of a set; see Ackerman's algebra notes, definition 0.4.1.1.

For each $g$, the function $x\mapsto g\cdot x$ is a bijection: its inverse is $x\mapsto g^{-1}\cdot x$, using the inverse element of $g$. Equivalently, a group action is a group homomorphism $\rho :G\to \mathrm{Sym}(X)$ into the symmetric group of all permutations of $X$, with $\rho (g)(x)=g\cdot x$ and $\rho (gh)=\rho (g)\circ \rho (h)$. See MIT's group-action notes, definition 23.1.

For a concrete group action, put the vertices of a polygon of a square at $A=(1,1)$, $B=(-1,1)$, $C=(-1,-1)$ and $D=(1,-1)$. Let $r(x,y)=(-y,x)$ be a counterclockwise quarter-turn rotation and let $s(x,y)=(y,x)$ be a reflection. Their eight permutations of the vertices of a polygon form the dihedral groupHere $D_4$ means the symmetry group of a four-sided polygon, with group order eight; the alternative convention calls this same group $D_8$. Products apply the rightmost operation first. For example, $s\cdot B=D$ and $(rs)\cdot B=r\cdot D=A$. The same group also has separate group actions on the sides of a polygon, the diagonals of a polygon, and the singleton set containing the centroid of the square. Specifying $X$ matters: changing what is moved changes the orbits of a group action and the stabilizer subgroups.

A group action need not be a faithful group action. For example, the cyclic group $C_4=\mathbb{Z}/4\mathbb{Z}$ acts on $X=\{0,1\}$ by $k\cdot x=x+k(\mathrm{mod}2)$, using modular arithmetic. Only the parity of $k$ matters: $0,2$ fix both points and $1,3$ exchange them. The associated group homomorphism $C_4\to \mathrm{Sym}(X)$ has group kernel $\{0,2\}$, of group order two. Either point has a group orbit of size two and a stabilizer subgroup of group order two, so the orbit-stabilizer theorem gives $4=2\cdot 2$.

The useful questions are which positions can be exchanged, and which transformations survive after imposing a constraint such as keeping one marked point fixed. An orbit of a group action answers the first question; a stabilizer subgroup answers the second. This turns a list of transformations into a count of distinct configurations, and explains why several transformations can produce the same configuration.

A half-plane is a half-space in a two-dimensional Euclidean space, consisting of points on one side of a line. For distinct centroids, each of the two Voronoi cells defined by their Euclidean distances is a closed half-plane bounded by their perpendicular bisector, with an explicit convention needed to assign boundary ties.

For a nondegenerate $d$-dimensional simplex $S$ with vertices $v_0,\dots ,v_d$ and an affine function $f$, the integral equals the volume times the value at the centroid $c=(v_0+\cdots +v_d)/(d+1)$. To prove this, write $x={\sum }_i{\lambda }_i(x)v_i$ in a barycentric coordinate system. On the standard simplex, symmetry permuting the vertices gives equal integrals of all ${\lambda }_i$; since ${\sum }_i{\lambda }_i=1$, each is $\mathrm{vol}(S)/(d+1)$. An invertible affine map transfers this result to any simplex. Affineness gives $f(x)={\sum }_i{\lambda }_i(x)f(v_i)$, so linearity of integration proves the formula. For a tetrahedron the factor is one quarter. The rule is exact for affine integrands; a nonlinear integrand generally needs additional moments.

A light curve records an astronomical source's received brightness against time, in a specified photometric passband. Its measured form is a time series of timestamps, brightness estimates, and uncertainties; a smooth $F(t)$ is a model of that record. Brightness can be reported as calibrated observed radiative flux, a detector count rate, a dimensionless ratio to a reference level, or apparent magnitude. These quantities require different units and calibration conventions. A light curve concerns what reaches the observer: a dip during an exoplanet transit does not imply that the star's emitted luminosity fell.

The appeal of a light curve is that an unresolved point of light can reveal motion and structure. In exoplanet transit photometry, repeated dips constrain the orbital period; under the uniform-disc, dark-planet, full-overlap assumptions, the exoplanet transit depth is the blocked area fraction $(R_p/R_{\star }{)}^2$. A toy one-percent dip therefore corresponds to a radius ratio of $0.10$. Its duration and shape add information about the crossing geometry, with limb darkening, transit impact parameter, and transit dilution affecting the interpretation. See exoplanet transit photometry for the geometry. The first figure instead follows the measurement: an instrumental trend and sky contribution can obscure a small physical dip.

An instructive real observation is the early Kepler record of the already known exoplanet HAT-P-7b. Welsh and colleagues analysed 33.5 days from 13 May to 15 June 2009, containing 15 complete exoplanet transits at an orbital period of about $2.20473$ days. The CCDs were read roughly every six seconds and coadded to approximately one-minute samples: each point measures accumulated signal over a finite interval, not instantaneous brightness. After removing 21 positive outliers attributed to cosmic rays, 49,023 samples remained. The authors estimated a scatter of about $150$ parts per million per one-minute sample. A gradual $0.13\%$ count decrease, suspected to arise from focus drift with a fixed extraction aperture, was removed with a cubic polynomial fitted while masking the exoplanet transits. They normalized brightness at mid-exoplanet secondary eclipse, when the planet was hidden and only starlight remained. ^[ref].

What does the instrument actually measure? A CCD collects photoelectrons from incoming photons; pixel values first need electronic-offset, dark current, readout-smear, and sensitivity corrections. ^[ref]. Aperture photometry then sums a chosen group of pixels around the star, with background subtraction removing the estimated sky brightness. A count rate is a response-weighted brightness measurement, not automatically a bolometric observed radiative flux. Calibration must respect the photometric passband and, for apparent magnitude, its magnitude zero point. Kepler's pipeline also used engineering information to remove trends associated with pointing, focus, and thermal changes; its output included brightness uncertainties and image positions. ^[ref].

Calibration and interpretation need separate checks. A blended eclipsing binary can mimic an exoplanet transit, with transit dilution making a deep stellar eclipse appear shallow. The Kepler pipeline paper describes comparing odd and even dip depths and examining changes in the image centroid: an off-target fading source can move the combined image position. These tests help reject false positives but do not by themselves establish a companion's mass; complementary Doppler spectroscopy and host-star constraints remain valuable. ^[ref].

Even a real exoplanet's signal can have several contributors. Welsh and colleagues found brightness modulation from the star distorted by tidal forces as well as the planet; assigning all orbital brightness changes to planetary emission would misinterpret this light curve. ^[ref]. Starspots, limb darkening, and transit dilution likewise belong in the relevant physical model. A flexible trend fitted through a dip could absorb that dip; the masked fit in the real example illustrates why the fitting procedure is part of the measurement. Exposure-averaged light curves and phase folding of a light curve make two further limitations explicit.

The descendant of a face with cyclic vertices $p_0,\dots ,p_{n-1}$ has vertices $p_i^{\prime }=(p_i+p_{i+1})/2$. Their arithmetic mean is unchanged. On the cyclic Fourier basis, a mode of frequency $j$ is multiplied by $(1+e^{2\pi ij/n})/2$. Every nonconstant mode has modulus less than one, so the descendant face shrinks to the original vertex centroid. This puts that centroid on the subdivision limit, without requiring the original face to be planar.

The midpoint of the line segment joining two points with position vectors $a,b$ is the point $m=(a+b)/2$. Its displacements from the endpoints are $(b-a)/2$ and $(a-b)/2$, so it lies on the segment and is equidistant from its endpoints. It is also the centroid of the two endpoints.

For two distinct centroids in a Euclidean space, equal squared Euclidean distances expand to the displayed affine hyperplane. It is their perpendicular bisector; each closed half-space consists of points at least as close to the centroid on that side. This proves weak linear separability after a two-centroid nearest-assignment step, allowing ties on the boundary. Coincident centroids are an essential exception: every point ties and the distance comparison supplies no nonzero normal vector.

The neutral axis is the line in the cross-section of an elastic beam where longitudinal bending strain vanishes. For a homogeneous beam in pure bending, zero resultant longitudinal force gives ${\int }_{A}zdA=0$, so the neutral axis passes through the centroid. Fibres on its opposite sides are stretched and compressed. The second moment of area about this axis controls the relation between bending moment and curvature.

For a group action of a group $G$ on a set $X$, the orbit of a group action of $x\in X$ is the subset of all positions reachable from $x$:Two points belong to the same group orbit exactly when one can be moved to the other by a group element. This is an equivalence relation: the identity element gives $x\sim x$, the inverse element of a transformation reverses a move, and a product combines two moves. Hence the group orbits form a partition of a set of $X$ into equivalence classes. A transitive group action on a nonempty set has just one group orbit. See Ackerman's algebra notes, lemma 0.5.1.3, for the orbit partition.

For the square and the eight-element dihedral group $D_4$ defined under group action, direct enumeration gives the following separate group orbits:

These counts depend on the chosen group action, not just the group. An orbit of a group action describes all configurations equivalent under the permitted transformations. For ordered pairs under the symmetric group, equality of the two entries is preserved, separating the two group orbits in equality-pattern orbits of ordered pairs; the orbit-stabilizer theorem computes their sizes when the symmetric group is $S_3$.

For a group action of any group $G$ on a set $X$, the cardinality of the orbit of a group action through $x$ is the index of a subgroup of its stabilizer subgroup:In particular, if $G$ is a finite group,For a fixed finite group, a larger stabilizer subgroup means a smaller group orbit: more transformations repeat the same image of $x$. The set $X$ itself need not be finite.

The underlying bijection uses left cosets, without requiring $G_x$ to be a normal subgroup:To check it briefly, replacing $g$ by $gh$ with $h\in G_x$ preserves $g\cdot x$, so $\Phi$ is well-defined. Every member of the group orbit has this form. Finally, $g\cdot x=k\cdot x$ implies $k^{-1}g\in G_x$, hence $g{G}_x=k{G}_x$, proving the bijection. This is the construction in MIT's group-action notes, lemma 23.5. For infinite groups, the bijection and the cardinality statement $|Gx|=[G:G_x]$ still hold; division of infinite cardinalities does not supply the finite formula.

For the eight-element dihedral group $D_4$ acting on the square described under group action, moving a marked vertex of a polygon gives $8=4\cdot 2$. Moving a marked side of a polygon also gives $8=4\cdot 2$; moving one unoriented diagonal of a polygon gives $8=2\cdot 4$; moving the centroid gives $8=1\cdot 8$. These are four different group actions, each with one group orbit. In the first group action, the four left cosets of $H=G_A=\{e,s\}$ areThey correspond respectively to $A,B,C,D$. Counting all eight transformations would count each position of the marked vertex of a polygon twice.

For a second computed example, let the symmetric group $S_3$ act on the Cartesian product $X=\{1,2,3{\}}^2$ by $\sigma \cdot (i,j)=(\sigma (i),\sigma (j))$. There are two orbits of a group action:andThe first stabilizer subgroup can exchange $2$ and $3$ while fixing $1$; the second must fix both $1$ and $2$, so also fixes $3$. The orbit-stabilizer theorem gives $6=3\cdot 2$ and $6=6\cdot 1$, and the two group orbit sizes account for all nine ordered pairs. The general reason for exactly these two group orbits is equality-pattern orbits of ordered pairs.

Combining the first two parts gives the centroid sum-of-squares identityThe added term is nonnegative and, since $n_k>0$, vanishes exactly at $z={\bar{x}}_k$. Thus the unique minimizing centroid in this step of K-means clustering isAn empty cluster needs a separate algorithmic convention because its sample mean is undefined and its objective is identically zero.