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For nonnegative integers $n,k$, the Stirling number of the second kindcounts the set partitions of an $n$-element finite set into exactly $k$ unlabelled blocks of a set partition. Set $S(0,0)=1$, $S(n,0)=0$ for $n>0$, and $S(n,k)=0$ for $k>n$. Also $S(n,1)=S(n,n)=1$ for $n\ge 1$. Grassl and Levin develop these counts through explicit set partitions and a recurrence relation. ^[ref]

Insert one distinguished element. For $n\ge 1$ and $1\le k\le n$, the Stirling numbers of the second kind obeyProof outline. Inspect the block of a set partition containing $n$. If it contains another element, removing $n$ leaves a set partition into $k$ blocks of a set partition; conversely, insert $n$ into any of those $k$ blocks of a set partition. If $n$ is alone, remove its singleton set and obtain a set partition into $k-1$ blocks of a set partition. The cases cannot overlap, and deletion reverses each construction. This explains the factor $k$ even though the blocks of a set partition have no names. ^[ref]

For $n=4,k=2$, start with the three two-block set partitions of $\{1,2,3\}$. Inserting $4$ into either block of a set partition gives six possibilities. The remaining possibility is $\{\{1,2,3\},\{4\}\}$, obtained by making $4$ a singleton set. Hence $S(4,2)=2\cdot 3+1=7$. The seven set partitions in the diagram comprise four with sizes $3+1$ and three with sizes $2+2$; this grouping by sizes independently gives $4+3=7$.

Three different counting questions. The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ selects one $k$-element subset. The Stirling number of the second kind $S(n,k)$ groups all $n$ distinct elements into $k$ nonempty unnamed blocks of a set partition. The power $k^n$ counts all functions to a specified $k$-element finite set, allowing some destination labels to go unused. At $n=4,k=2$, these counts are respectively $6$, $7$, and $16$. Choosing two people, making two unnamed nonempty teams, and assigning each person a red or blue label are different tasks.

When the destinations have labels. For fixed finite sets $X,Y$ of sizes $n,k$, respectively, the number of surjective functions $X\to Y$ isHere every element of $Y$ must be used. Each surjective function gives a set partition of $X$ into its nonempty preimages $f^{-1}(\{y\})$. Conversely, assign the $k$ distinct destination labels to the $k$ blocks of a set partition in $k!$ ways, where $k!$ is a factorial. For two labelled teams of four people this gives $2!\cdot 7=14$: of the $2^4=16$ assignments, exactly two use only one label. Dividing arbitrary function counts by $k!$ fails in general because unused labels can be permuted without changing a function. For example, $3^4/3!=13.5$ cannot count set partitions. The empty-to-empty function is a surjection, consistent with $0!S(0,0)=1$. Stanley gives the bijection underlying this qualification.^[ref]

For example, the inclusion-exclusion principle counts the surjections from four elements to three labelled destinations as $3^4-3\cdot 2^4+3\cdot 1^4=36$. Thus $S(4,3)=36/3!=6$. An independent check chooses the only two-element block of a set partition in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways; the other two elements form singleton sets.

Why these numbers also occur in algebra. Classifying all functions to an $m$-element finite set by the size $k$ of their image givesFor a fixed set partition, the falling factorial $(m{)}_k$ counts assignments of distinct labels to its $k$ blocks of a set partition, and is zero when $k>m$. At $n=3,m=4$, this reads $64=1\cdot 4+3\cdot 12+1\cdot 24$. Replacing $m$ by $x$ yields a polynomial identity: both sides agree at every nonnegative integer. Thus the Stirling numbers of the second kind translate powers into falling factorials. Grouping by equal values turns a counting rule into an algebraic change of coordinates. Summing $S(n,k)$ without destination labels instead gives the Bell number.

A set partition $\mathcal{Q}$ refines $\mathcal{P}$ if every cell of $\mathcal{Q}$ is contained in a cell of $\mathcal{P}$. Thus $\mathcal{Q}$ retains every separation made by $\mathcal{P}$ and may make additional separations. Refinement of a set partition is reflexive and transitive. If each partition refines the other, their cells coincide, so refinement of a set partition is a partial order on partitions. For finite partitions the nonempty intersections of cells form their common refinement of a set partition. In graph regularity arguments, subdividing cells and treating discarded vertices as singletons preserves refinement of a set partition; merging discarded vertices into one cell need not. The refinement variance identity for regularity energy consequently proves that this singleton treatment cannot decrease the equitable regularity energy.

A block of a set partition is one of its nonempty subsets. Distinct blocks of a set partition have empty intersection, and their union is the original set. For $\pi =\{\{1,3\},\{2\}\}$, the blocks of a set partition are $\{1,3\}$ and the singleton set $\{2\}$. The indices in a written list of blocks of a set partition do not label or order the set partition.

For the nondimensional simple pendulum $\ddot{x}=-\sin x$, the separatrix has energy ${\dot{x}}^2/2+1-\cos x=2$. Starting at $x=0$ with velocity $2$, its time to angle $a<\pi$ is $\log (\sec (a/2)+\tan (a/2))$. The saddle at angle $\pi$ is approached only as time tends to infinity.

// Target: classical-mechanics.bigb

A probabilistic model assigning distinct keys independent bucket indices with a uniform distribution on a finite set. It supplies an expected value bound for separate chaining, not a guarantee for every deterministic hash function or workload.

For a neutral sample of size $n\ge 2$ under the infinite sites mutation model, the Watterson estimator is ${\hat{\theta }}_W=S/a_{n-1}$, where $a_{n-1}={\sum }_{i=1}^{n-1}1/i$. Since the coalescent has $ES=\theta a_{n-1}$, this estimator is unbiased. It matches the observed segregating-site count to its expected value, but need not maximize the exact segregating-site likelihood under the neutral coalescent.

The Maykut-Untersteiner model is a seasonally forced one-dimensional sea ice thermodynamics model with snow and saline ice layers. It couples radiative and turbulent surface fluxes, heat conduction, interior shortwave absorption, brine-dependent effective heat capacity and conductivity, snowfall, and a basal Stefan condition driven by prescribed ocean heat flux. Its annual sea-ice thermodynamic equilibrium is a periodic cycle with zero net yearly thickness change. The snow insulation and summer ice protection mechanisms compete, so additional snow need not reduce annual equilibrium thickness. The transient temperature profile need not be linear.

When seawater freezes, most salt remains in liquid, increasing salinity and density and potentially driving convection.

The water-equivalent thickness expresses an ice mass per area as a depth of water at a declared reference density. With $m_i={\rho }_i{h}_i$ per area, $h_w=m_i/{\rho }_{w,\mathrm{ref}}$. Ice at $900\mathrm{kg}{\mathrm{m}}^{-3}$ and thickness $2.44\mathrm{m}$ has about $2.20\mathrm{m}$ water equivalent at a fresh-water reference density of $1000\mathrm{kg}{\mathrm{m}}^{-3}$. The depth of seawater consumed at $1025\mathrm{kg}{\mathrm{m}}^{-3}$ is instead about $2.14\mathrm{m}$. A reference depth must not replace ice thickness in an ice-volume calculation or the liquid mass removed in brine rejection.

A freezing degree-day accumulates air temperature below a declared reference freezing temperature, with time in days and temperature difference in degrees Celsius or kelvins. The reference temperature and averaging convention must match the empirical sea ice thermodynamics formula using it. Warming by a constant $\Delta T$ subtracts $\Delta T$ times the number of days only if those days remain below the reference threshold.

Melting a basal ice thickness $\Delta h$ consumes energy ${\rho }_i{L}_f\Delta h$ per unit area, ignoring sensible warming and salinity corrections. The corresponding time-averaged heat flux is this energy divided by $\Delta t$. The actual ocean heat flux also includes heat conducted into the ice.

// Target: geophysics.bigb

Every spherically symmetric vacuum solution of the Einstein field equation is locally a portion of Schwarzschild spacetime. In particular, a spherically symmetric body's exterior field depends only on its total mass.

Geodetic precession is the rotation of a parallel-transported gyroscope relative to the orbital radial direction in curved spacetime. For a circular Schwarzschild orbit its exact precession per revolution is $2\pi [1-\sqrt{1-3M/R}]$.

Inside the future black-hole region, the radius decreases along every future-directed timelike path. Its unit tangent obeys $|dr/ds|\ge \sqrt{2M/r-1}$ from the metric normalization, independently of radial acceleration or angular motion. Integrating from $r_0<2M$ to zero gives the displayed finite upper bound. The statement concerns future continuation within the black-hole region; reversing time in the white-hole region reverses the radial direction.

For radial timelike geodesic fall released from rest at $r_0>2M$, the Killing energy is $E=\sqrt{1-2M/r_0}$ and the proper time radial equation is $(dr/ds{)}^2=2M/r-2M/r_0$. The displayed parameterization integrates its inward branch, with $s=0$ at release. The Schwarzschild event horizon is reached at ${\eta }_H=\arccos (4M/r_0-1)$ in finite proper time, even though Schwarzschild time diverges.

// Target: general-relativity.bigb

A holomorphic self-map of the unit disc contracts its pseudohyperbolic distance. Conjugating by Möbius transformations that send one chosen source point and its image to zero reduces the statement to the Schwarz lemma. In the complex upper half-plane this givesIf $f(iY)=iY+o(Y)$ at infinity, multiply the equivalent inequality between one minus the squared distances by $Y$ and let $Y\to \infty$. It yields $\mathrm{Im}f(z)\ge \mathrm{Im}z$. Applied to the inverse mapping-out function of a compact H-hull, this proves the nonnegativity of its half-plane capacity.

If a holomorphic function $F$ on the unit disc has positive real part, the Cayley transform between the half-plane and disk $\varphi =(F-F(0))/(F+\overline{F(0)})$ maps the disc into itself and vanishes at zero. The Schwarz lemma gives the displayed derivative bound. Equality holds exactly when $\varphi (z)=\omega z$ with $|\omega |=1$, equivalently $F(z)=i\mathrm{Im}F(0)+\mathrm{Re}F(0)(1+\omega z)/(1-\omega z)$.

For a zero-free holomorphic function $f:\mathbb{D}\to \mathbb{D}$, put $c=f(0)$ and $\alpha =-\log |c|$. A holomorphic logarithm makes $-\log f$ have positive real part. The Carathéodory derivative bound for the right half-plane yields the displayed sharp bound. Equality functions are $f(z)=c\exp [-2\alpha \omega z/(1-\omega z)]$, $|\omega |=1$. The Harnack inequality on the unit disk also yields the sharp pointwise bounds $|c{|}^{(1+r)/(1-r)}\le |f(w)|\le |c{|}^{(1-r)/(1+r)}$ for $r=|w|$. Aligning $\omega w=\pm r$ attains the two endpoints.

Partition jobs between machines and choose an ordering on each machine. Machine $r$ has load $C_r$ equal to its processing times, initial setup and internal changeovers; the completion time of all jobs is the maximum of these loads. Minimize a variable $C_{\max }$ subject to $C_r\le C_{\max }$ for every machine. Branch and bound can combine machine-assignment decisions with route bounds; heuristic moves relocate jobs between machines, swap jobs, and reorder blocks. Minimizing the sum of route costs alone does not generally minimize the makespan.

Cosmic expansion is the increase of physical distances between comoving observers as the scale factor grows. Its instantaneous fractional rate is the Hubble parameter.