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Expanding the squared Euclidean distances givesFor distinct centroids, chooseA point assigned to $C_1$ has the displayed distance difference at most zero and therefore $w^{\mathsf{T}}x+b\ge 0$; assignment to $C_2$ gives the reverse inequality. The separating hyperplane is the perpendicular bisector of the line segment joining the two centroids. Points tied between them lie on that hyperplane. This is the nearest-centroid decision hyperplane.

The requested nonzero $w$ needs the additional assumption $z_1\ne z_2$. It does not follow from the stated algorithm alone. In one dimension, take $C_1=\{-1,1\}$ and $C_2=\{0\}$, whose centroids are both zero. A tie rule that retains previous assignments can leave this configuration after an iteration. The inequalities at $-1,1$ imply $b\ge |w|$, while the inequality at zero implies $b\le 0$, forcing $w=0$. Thus the unconditional nonzero-separator claim has this degenerate exception.

The boundaries intersect where $x^2=2x+8$, giving $x=-2$ and $x=4$. Between these values, the straight line lies above the parabola. ThereforeEvaluation givesA useful independent check uses the region's symmetry about $x=1$: its vertical width is $9-(x-1{)}^2$. Its area is ${\int }_{-3}^3(9-u^2)du=36$, and its centroid has $x$ coordinate one, so its first moment $\iint xdA$ is again $36$.

Move the origin to the centroid and write the four vertex vectors as $r_1,r_2,r_3,r_4$. Their sum is zero. The midpoints of the first opposite-edge pair have vectors $(r_1+r_2)/2$ and $(r_3+r_4)/2=-(r_1+r_2)/2$, so their distances from the centroid are equal. The same argument applies to the other two opposite-edge pairs.

Choose the three midpoint distances as $u=|r_1+r_2|/2$, $v=|r_1+r_3|/2$, $w=|r_1+r_4|/2$. Expanding the dot products,Identifying these four norms with the vertex distances proves the tetrahedron centroid-to-midpoint sum of squares:The proof uses no orthogonality or regularity assumption on the tetrahedron.

This is mid-edge subdivision, rather than a vertex-retaining refinement. Use the usual closed two-manifold mesh setting: each edge has two incident facets and each vertex has a cyclic fan. For boundary meshes, a boundary refinement rule has to be specified separately. The relevant generic-position condition excludes collapsed local tangent directions and singular regular patches.

Topology and the manifold property. Regard the abstract mesh as a cell decomposition of a surface. Join the edge midpoints in each facet. The inner polygon is the new face associated with that facet; the corner regions around an old vertex assemble into the face associated with that vertex. At an old edge midpoint, the four adjacent corner connections form a cyclic link. Thus every new interior vertex has valence four, every new edge has two incident faces, and the refined cell complex is homeomorphic to the original surface. If the original counts are $V,E,F$, the new counts are $E,2E,V+F$, preserving $V-E+F$. The cyclic-link argument, rather than Euler characteristic alone, verifies the local manifold condition. Convergence and local regularity, justified below, transfer this abstract surface structure to the limit.

Convex hull. Each newly created point is an average of two previous points. By induction it is a convex combination of original vertices. The same is true of points of any triangulation of the refined facets, and taking limits keeps them in the closed convex hull. Hence the entire limit stays in that hull, not necessarily inside the original nonconvex enclosed solid.

Facet centroids. Follow the descendant inner face of an original $n$-sided facet, with cyclic vertices $p_i$. Its vertices satisfy $p_i^{(r+1)}=(p_i^{(r)}+p_{i+1}^{(r)})/2$. Summing proves mid-edge facet centroid invariance:For a cyclic Fourier mode $e^{2\pi iji/n}$ the multiplier is $(1+e^{2\pi ij/n})/2$, whose modulus is $|\cos (\pi j/n)|<1$ for $j\ne 0$. Every mean-zero mode therefore decays, so all descendant face vertices converge to $\overline{p}$. Consequently the original vertex centroid belongs to the limit surface. No planarity assumption is needed.

Quadratic triangular pieces. On a regular quadrilateral mesh, apply two midpoint steps. Each new point combines three old neighboring controls with weights $1/2,1/4,1/4$. For example, around a face with cyclic corners $p_{00},p_{10},p_{11},p_{01}$, the four new points areThis is the binary refinement mask obtained by expandingwith the translation chosen to place the refined grid at quarter offsets. The limit is therefore represented by shifts of the quadratic four-direction box spline with directions $e_1,e_2,e_1+e_2,e_1-e_2$. Projection of a four-cube onto the plane has piecewise total degree $4-2=2$, and its knot grid is triangular. Coordinatewise linear combinations preserve that degree, giving parametric quadratic triangle patches. Around a nonregular center there are infinitely many shrinking rings of such patches; the isolated center is their limit, not generally part of a finite polynomial patch representation.

Only finitely many initially possible smoothness exceptions. Every vertex has valence four after the first step. Only the descendants of original facets and the faces created around original vertices can remain nonquadrilateral. All subsequently created faces around four-valent vertices are quadrilaterals. Thus every point except the limiting centers of those persistent nonquadrilateral faces eventually lies in a regular patch neighborhood. The number of possible exceptional centers is at most $V+F$. For the regular box spline, any three distinct directions still span the plane, so three directions must be removed to make the remainder nonspanning. The box spline continuity criterion gives $C^{3-2}=C^1$ across its triangle boundaries. This proves $C^1$ continuity away from at most those finitely many centers.

Removing the apparent exceptions. Center a finite neighborhood of an $n$-sided exceptional face, and analyze a full two-step subdivision matrix so the mesh orientation is restored. Cyclic Fourier basis modes diagonalize its sector symmetry. The face mode has eigenvalueThe local ring blocks also contribute eigenvalues $1/4$ and zero. For $n>3$, the constant mode is the single eigenvalue one, and the two first-frequency tangent modes dominate all nonconstant remaining modes. Construct their characteristic map of a subdivision surface; its quadratic derivative pieces must have a nonzero consistently oriented Jacobian, and its fundamental sector must be injective. These checks, not the eigenvalue ordering alone, give a regular chart and a continuously limiting tangent plane for generic data.

The case $n=3$ needs a separate Jordan analysis: $\lambda =1/4$ is eightfold, so the usual semisimple double-eigenvalue test does not apply. Its first-frequency Jordan chains supply the leading tangent behavior, while the other contributions become relatively negligible. Regularity and injectivity of that generalized characteristic map complete the argument. These concrete spectral and characteristic-map checks are established for this scheme in Peters and Reif, Sections 3 and Appendix. Thus the generic limit is geometrically $C^1$ also at the exceptional centers; compatible local charts need not preserve the original uniform refinement parameterization.

The manifold statement concerns the abstract limit surface and its regular local parametrizations. It does not, from combinatorics and generic tangent data alone, prove global injectivity of the image in three-dimensional space. A particular model may require an additional global self-intersection test before claiming that its image is an embedded manifold.

Cluster analysis groups observations without given class labels. It starts by deciding what similarity means, usually through a dissimilarity matrix. Euclidean distance is natural for comparable continuous coordinates; a Mahalanobis distance can account for differing scales and correlations when a meaningful nonsingular covariance estimate is available. Scaling and variable selection can change the groups. Binary, ordinal or categorical data may need a different dissimilarity, rather than arbitrary numerical codes and Euclidean distance.

Agglomerative hierarchical clustering begins with singleton groups, repeatedly merges the pair with smallest intergroup dissimilarity, and records the merges in a dendrogram. The height represents the algorithm's merge criterion. Cutting the dendrogram gives partitions at different resolutions, and merges cannot subsequently be reversed. In single-linkage clustering, the criterion is the nearest cross-pair distance; this detects chains and nonconvex groups, but a thin bridge can connect otherwise separated clouds. Complete-linkage clustering uses the farthest cross-pair distance and tends to produce compact groups, at the cost of sensitivity to extreme points. Average-linkage clustering uses the mean cross-pair distance, lying between these two criteria for a given pair of clusters.

For squared Euclidean geometry, Ward clustering merges the pair that least increases the within-group sum of squares. Expanding deviations about the merged mean givesThis favors low within-group dispersion, but interpreting it as a sum-of-squares increase requires Euclidean data. Different software may plot a rescaled height, so the vertical units of a dendrogram from Ward clustering must be stated.

A nonhierarchical alternative is k-means clustering. For a prescribed $K$, minimizeAt fixed assignments, expanding squared distances or differentiating shows that the minimizing $m_k$ is the group mean. At fixed centroids, each point is assigned to a nearest centroid. Alternating these operations does not increase $W_K$; with a consistent tie rule and nonempty groups it reaches a locally stable partition. Initialization matters, so use multiple starts, and give an explicit rule for empty groups. It favors roughly spherical clusters and is sensitive to outliers and scaling. It cannot be expected to reproduce a nonconvex single-linkage partition.

The sketch displays a constructed point configuration and the complete-linkage merges computed from its actual Euclidean distances. A cut between the within-pair and between-pair merge heights yields three groups. Such visual separation is useful evidence, not proof of a uniquely correct number of populations. Select a resolution using subject knowledge, stability under resampling, separation diagnostics or the change of $W_K$ with $K$. Clustering always gives a partition even for unstructured data, and different distance or linkage choices can yield different sensible summaries. Clustering is an exploratory construction whose conclusions depend on the geometry and the chosen resolution.

For $y\ne 0$, the squared Euclidean norm of the orthogonal projection residual is nonnegative:This proves the Cauchy-Schwarz inequality $|x\cdot y|\le |x||y|$. Equality holds exactly when $x$ is a scalar multiple of $y$. If $y=0$, equality is automatic. Equivalently, equality means the two vectors are linearly dependent, including the cases with a zero vector.

For a point $C$ equidistant from the coordinate vertices,must be independent of $i$. Thus all $C_i$ are equal, and their sum is one. The centroid is thereforeFor every $x$ in the base probability simplex, the Cauchy-Schwarz inequality gives $1=(1,\dots ,1)\cdot x\le \sqrt{n}|x|$. Equality forces the same coordinates as $C$. Thus $C$ is the unique closest point of a probability simplex to the origin.

The angle $\alpha$ between $OC$ and any edge from the origin to $e_i$ satisfiesConsequently

Use the corrected decompositionThe hint's minus sign before $({\mu }_j-w_j)$ is a typographical error. Squaring, summing, and using ${\sum }_i(y_{ij}-{\mu }_j)=0$ for each coordinate gives the centroid sum-of-squares identitySince $m>0$ and $w\ne \mu$, the final term is strictly positive. HenceThus the centroid uniquely minimizes the sum of squared Euclidean distances to the observations.

Write $x_i\in {\mathbb{R}}^p$ for row $i$ of the data matrix. The k-means clustering objective is the within-cluster sum of squared Euclidean distances:The nonempty $C_k$ form a partition of a set of observation indices. For fixed clusters, part (a) shows that the best $m_k$ is their centroid.

Initialize distinct centres, for example by choosing $K\le n$ distinct observations. Then repeat the following steps:

The empty-cluster repair is possible for $K\le n$ and cannot increase $J$: the transferred observation now has squared distance zero, while all other squared distances stay unchanged before the centre update.

Assignment cannot increase $J$, and any changed assignment strictly decreases it under the tie rule. Updating centres cannot increase $J$ by the centroid sum-of-squares identity. After an update, the centres are determined by the partition of a set. There are only finitely many possible assignments. Unless this is the initial round, any nonterminal round starts with an assignment that strictly reduces $J$; repairing empty clusters and updating cannot undo that decrease. Thus a partition cannot recur, proving finite termination of K-means clustering. The final result need not be a global minimum; different initial centres can lead to different final clusterings.

Write the centroid as $\overline{\mathbf{r}}=\overline{x}\mathbf{i}+\overline{y}\mathbf{j}$. The region has reflection symmetry in the $y$-axis: replacing $\theta$ by $\pi -\theta$ changes $x$ to $-x$ and preserves $y$ and $dA$. Thus symmetry in integration givesFor the other first moment of area, use $y=r\sin \theta$ in polar coordinates:Here the last integral has the value allowed in the paper. Dividing the first moment of area by the area found in i givesThe positive vertical coordinate agrees with the fact that the entire region lies above the $x$-axis.

For observations ${\mathbf{x}}_i\in {\mathbb{R}}^p$, K-means clustering seeks nonempty clusters in cluster analysis $C_1,\dots ,C_K$ and centroids ${\mathbf{m}}_k$ minimizingAt fixed centroids, assigning each point to a nearest centroid minimizes its squared Euclidean distance. At fixed assignments, the centroid sum-of-squares identity shows that the sample mean uniquely minimizes each group's contribution. These alternating steps cannot increase $W$.

For finite termination of K-means clustering, more than a lower bound on $W$ is needed. Keep an assignment unchanged when nearest centroids tie, and stop when assignments no longer change. Every change then strictly decreases $W$. After updating the centroids, $W$ is determined by the partition of a set of $n$ indices; there are only finitely many choices for that partition of a set, so no strictly improving partition of a set can recur. Empty clusters in cluster analysis require a convention, such as moving a point from a nonsingleton cluster in cluster analysis to become its own centroid, which cannot increase $W$. With such a consistent convention the algorithm terminates, although it may reach only a local minimum.

With two distinct centroids, K-means clustering assigns points on opposite sides of their nearest-centroid decision hyperplane, which in two dimensions is a straight line. This constraint explains the typical outcomes in the three source plots; exact assignments depend on initialization and the realized points.

Dataset A: one cluster in cluster analysis contains most of the horizontal cloud and the bottom of the vertical cloud; the other contains the upper part of the vertical cloud. The two visually elongated groups are poorly matched to squared Euclidean distance from their centroids. Points at the lower end of the vertical group can be closer to the horizontal group's centroid, so the two original shapes need not be recovered intact.

Dataset B: a roughly straight cut divides the ring and generally the central cloud between the two clusters in cluster analysis. A Voronoi cell for either centroid is a half-plane; it cannot select a surrounding ring while excluding its center. The cut's direction can vary because the geometry is approximately rotationally symmetric.

Dataset C: the two dense lower clouds typically receive different centroids. The much smaller upper cloud is absorbed into one or divided between both clusters in cluster analysis, according to the nearly vertical separating line. With $K=2$, the objective usually benefits more from separating the two large clouds than from spending one centroid on the sparse third group. These are typical qualitative outcomes, rather than guarantees about every initialization.

In the $x,z$ plane, the embankment lies above the parabola $z=H{x}^2/L^2$ and below $z=H$. The bore has circular cross-sectionIt touches the valley floor at the origin. To check that this entire disk lies inside the projected embankment, its equation gives $x^2\le Dz-z^2\le Dz$. Since $HD/L^2<1$, it follows that $H{x}^2/L^2\le z$, with strict inequality away from the bottom contact. Also $z\le D<H$. Thus no circular segment needs to be subtracted from the tunnel section.

For each $(x,z)$ in the disk, the length of bore inside the material is the local width $2b(2-z/H)$. The required volume integral is thereforeThe disk has area $A=\pi D^2/4$ and centroid height $D/2$, so ${\iint }_{\mathcal{D}}zdxdz=AD/2$. Applying the volume of a solid with affine thickness formula givesThe width varies along the circular section; multiplying its area by the top width alone would miss this variation.

The divergence isFor the tetrahedral domain, convenient integration bounds areConsequently its volume integral isPutting $h=2(1-x-y)$, the inner integral is $(2y+6)h+h^2=(16-4x)(1-x-y)$. ThereforeThusThere is an independent centroid check. The vertices are $(0,0,0),(1,0,0),(0,1,0),(0,0,2)$, giving volume $1/3$ and centroid $(1/4,1/4,1/2)$. An affine function averages to its value at the centroid, so $I=(1/3)(6+2/4+2/2)=5/2$. This is the integral of an affine function over a simplex.

For uniform density, the centroid is the volume average of position and coincides with the centre of mass:A point fixed in the body has ${\mathbf{r}}_{OQ}={\sum }_i{q}_i{\mathbf{e}}_i$ with time-independent body coordinates $q_i$. Since $O$ is fixed, part (a) givesThe centroid is fixed relative to the body as well, so ${\mathbf{v}}_G=\mathbf{\omega }\times {\mathbf{r}}_{OG}$. Using ${\mathbf{r}}_{OQ}={\mathbf{r}}_{OG}+{\mathbf{r}}_{GQ}$ therefore gives

Write $r=(x_1,x_2,x_3{)}^T$ for position relative to the centroid. The inertia tensor there isIn particular its diagonal entries integrate the squares of distances from the corresponding axes, and its off-diagonal entries are the negative products of coordinates.

Insert ${\mathbf{v}}_Q={\mathbf{v}}_G+\mathbf{\omega }\times r$ into $T=\frac{1}{2}{\int }_A\rho |{\mathbf{v}}_Q{|}^{2}dV$. The cross term vanishes because ${\int }_A\rho rdV=0$. Using $|\mathbf{\omega }\times r{|}^2=|\mathbf{\omega }{|}^2|r{|}^2-(\mathbf{\omega }\cdot r{)}^2$ givesIf $a$ is the vector from $G$ to $O$, positions relative to $O$ are $r-a$. Expanding the defining tensor integral with this substitution again removes the terms linear in $r$. Thus the tensor parallel axis theorem isSince $O$ is fixed, ${\mathbf{v}}_G=\mathbf{\omega }\times (-a)$, so $m|{\mathbf{v}}_G{|}^2=m{\mathbf{\omega }}^T(|a{|}^{2}I-a{a}^T)\mathbf{\omega }$. Substitution into the energy decomposition yields

Choose the $z$-axis along the cylinder and the origin at its midpoint, so $x^2+y^2\le b^2$ and $-a\le z\le a$. Reflection symmetry puts the centroid at the origin and makes all products of inertia zero. The mass isCylindrical integration gives ${\int }_A\rho x^{2}dV={\int }_A\rho y^{2}dV=m{b}^2/4$, while the uniform height coordinate has ${\int }_A\rho z^{2}dV=m{a}^2/3$. ConsequentlyThe two equal transverse entries reflect the rotational symmetry about the cylinder axis.

By symmetry, the resultant Coulomb force on each charge points along the line from the centroid to its vertex. Their accelerations and subsequent speeds are equal, so the centroid remains fixed and the equilateral triangle expands without rotating. Each charge follows a straight radial trajectory; the separation grows and the repulsive force decreases.

For example, if the instantaneous side is $s$ and the centroid-to-charge distance is $r=s/\sqrt{3}$, the two pair forces have a $60^{\circ }$ angle between them. Their resultant is $\sqrt{3}{k}_e{q}^2/s^2=k_e{q}^2/(\sqrt{3}{r}^2)$, along the outward radius. This confirms that the symmetric shape persists.

Assuming release from rest, nonrelativistic motion and negligible radiation losses, energy conservation givesThe three velocity vectors sum to zero, consistent with momentum conservation. Their magnitudes approach the displayed value as the remaining electrostatic potential energy tends to zero.

Take $\zeta$ as signed distance from the neutral axis within the beam cross-section. Bending into a large radius $R$ changes the length of a fibre proportionally to $\zeta /R$, so linear elasticity gives $\sigma =E\zeta /R$, up to the curvature sign convention. The moment magnitude isHere $I$ is the second moment of area about the neutral axis, with units of length to the fourth power, not the mass moment of inertia. A homogeneous cross-section has its neutral axis through the centroid under pure bending.

For the straight pin-ended beam, the compressive dead load produces bending moment $-Fy$ at transverse offset $y$. In the small-slope approximation, curvature is $y^{\prime \prime }$ with the corresponding sign convention, givingThe stated endpoints fix positions rather than slopes: this interpretation is consistent with the illustrated end forces and the later sinusoidal unstressed shape. Truly clamped endpoints would have a different threshold. A nonzero solution $y=A_n\sin (n\pi x/L)$ requires $F_n=n^2{\pi }^{2}EI/L^2$. Thus the first Euler load is $F_c={\pi }^{2}EI/L^2$. Below it the perfect straight state is stable, at it the first mode is marginal in linear theory, and above it the straight state is unstable. The linear equation selects thresholds but cannot set the amplitude or describe a finite postbuckled equilibrium.

For the initially curved compressed beam, elastic stress responds to the change from its stress-free curvature. With $y=y_0+\eta$, the moment is $EI(y^{\prime \prime }-y_0^{\prime \prime })=EI{\eta }^{\prime \prime }$, so force balance givesPut $q=\pi /L$ and seek the branch continuously connected to the unloaded shape. Since $y_0=Y\sin qx$, substitute $\eta =C\sin qx$ to obtain $-EI{q}^{2}C=-F(Y+C)$, orAway from eigenvalues, the homogeneous solution $A\sin (\sqrt{F/(EI)}x)+B\cos (\sqrt{F/(EI)}x)$ vanishes under the two end conditions, so this particular solution is the unique linear equilibrium. For $0\le F<F_c$ and $Y>0$, the maximum additional displacement is ${\eta }_{\max }=YF/(F_c-F)$ and maximum total displacement is $y_{\max }=Y/(1-F/F_c)$, both at midspan.

This sinusoidal imperfection amplification of a column produces deflection at every positive load and strong amplification near $F_c$. The apparent divergence signals loss of the small-slope model before infinite displacement; nonlinear geometry is needed there. The formal continuation above the first threshold is not a stable physical loading branch. Both total and additional displacement are distinguished in the graph.

Archimedes' principle states that the resultant hydrostatic pressure force on a body equals the weight of the fluid displaced, directed opposite to gravity. Let $V_d$ be the displaced fluid region, with outward normal $\hat{\mathbf{n}}$ from the body into the fluid. Since the fluid pressure force is inward,Hydrostatic balance is $\nabla p={\rho }_f\mathbf{g}$, with downward gravitational acceleration $\mathbf{g}$. Thereforefor uniform fluid density and gravity. Uniform atmospheric pressure gives no net force on a closed surface. For a floating body, the same displaced-volume construction includes the free surface; any significant air buoyancy can be included as a second displaced fluid.

A freely floating body of mass $m$ satisfies $mg={\rho }_{f}g{V}_d$, so its displaced volume is $m/{\rho }_f$. A homogeneous body of density ${\rho }_b<{\rho }_f$ floats with submerged fraction ${\rho }_b/{\rho }_f$; if ${\rho }_b>{\rho }_f$, hydrostatic buoyancy alone cannot support it. The buoyant force acts through the centre of buoyancy, the centroid of the displaced volume for constant density, while weight acts through the centre of mass. Their relative positions determine rotational equilibrium and stability. For a fully submerged body, a centre of mass below the centre of buoyancy gives a restoring torque; a floating body's changing displaced shape additionally requires the metacentric height criterion. Surface tension, externally driven flow and contact forces can alter the simple force balance.

For a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.