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An elastic collision preserves total kinetic energy between the incoming and outgoing states. For isolated colliding bodies, momentum conservation also holds. The kinetic energy condition supplies information beyond momentum conservation alone. ^[ref]

Consider a stipulated one-dimensional example with rightward positive: $m_1=2\mathrm{kg}$, $m_2=1\mathrm{kg}$, incoming velocities $u_1=3m/s$ and $u_2=0$. The total momentum is $P=6\mathrm{kg}m/s$ and total kinetic energy is $K=9\mathrm{J}$. In SI numerical values, outgoing velocities $v_1,v_2$ satisfySubstituting $v_2=6-2v_1$ gives $3(v_1-1)(v_1-3)=0$. The root $v_1=3$, $v_2=0$ repeats the incoming state; the outgoing separating state is $v_1=1m/s$, $v_2=4m/s$. Its momenta add to $2+4=6\mathrm{kg}m/s$ and its kinetic energies add to $1+8=9\mathrm{J}$. The incoming and outgoing relative velocities both have magnitude $3m/s$, consistent with coefficient of restitution $e=1$.

The inelastic collision example below uses exactly these incoming data. Both outcomes preserve total momentum, while only the elastic collision preserves total kinetic energy. These are ideal numerical examples, not experimental measurements.

For chain complexes of free modules over a principal ideal domain $R$, the Künneth theorem gives a natural split short exact sequencealthough the splitting is not natural. Over a field the Tor term vanishes, and the cohomological cross product gives $H^{\ast }(X\times Y;k)\cong H^{\ast }(X;k){\otimes }_k{H}^{\ast }(Y;k)$ as graded rings under standard finiteness hypotheses.

The cup product is a natural bilinear operation $H^p(X;R)\times H^q(X;R)\to H^{p+q}(X;R)$. It is graded-commutative: $a⌣b=(-1{)}^{pq}b⌣a$.

The cap product pairs a homology class and a cohomology class to lower degree:

Compactly supported cohomology is the cohomology of cochains vanishing outside a compact subset. Equivalently,

This is the cohomological form of the universal coefficient theorem, with Ext and Hom terms rather than the homological tensor and Tor terms.

For an abelian coefficient group $G$, cohomology fits into a split short exact sequence

A limit point of a subset $A$ of a topological space is a point $x$ whose every neighbourhood meets $A\setminus \{x\}$. Thus $1$ is a limit point of $[0,1)$ in the real line, even though $1\notin [0,1)$. In contrast, an isolated point of $A$ belongs to its closure but is not a limit point of $A$. This distinction separates limit points from the more inclusive notion of an adherent point.

For $\beta =4$ and $a(0)=0$, the expanding branch isIt grows as $2\sqrt{H_{0}t}$ near zero and approaches $\sqrt{2}$ exponentially at late times.

For $\beta >4$, the turning polynomial has two positive roots and its smaller root lies below $2$. A universe expanding from $a=0$ reaches that root with $H=0$ and $\dot{H}<0$, then contracts to a finite-time Big Crunch.

If $X,Y$ are metric spaces, $Y$ is compact, and the graph of $f:X\to Y$ is closed, then $f$ is continuous. Any failure of sequential continuity gives a subsequence whose images stay away from the proposed limit; compactness produces a convergent image subsequence, and closedness of the graph forces its limit to be the correct value.

Integrating Maxwell–Boltzmann statistics over momentum for nonrelativistic particles of mass $m$, spin-state count $g_s$, and energy $m{c}^2+p^2/(2m)$ gives

A class $M$ is transitive when $x\in y\in M$ implies $x\in M$. Consequently, every member of a member of $M$ is available as an element of the structure $(M,\in )$.

A proper class is a class in set theory that is not a set in the ambient universe.

For the clamped-free beam under uniform load and endpoint force, the additional minimized internal energy isIts derivative is the endpoint displacement caused by the endpoint force:

At chemical equilibrium, a reversible chemical reaction has equal forward and reverse reaction rates, so its macroscopic composition stays constant. Individual molecules can keep changing identity: the reaction has not stopped. This dynamic balance is the definition described by the ^[ref]. Equal reaction rates also need not mean equal concentrations. A small population converting rapidly can balance a large population converting slowly.

This distinction gives two useful questions. Which composition is favored at the specified temperature and other constraints? Thermodynamics answers through chemical potentials and an equilibrium constant. How quickly does a prepared mixture approach that composition? Chemical kinetics answers through reaction rates. Knowing the destination does not determine the journey's duration. A composition that barely changes during observation may reflect extremely slow chemical kinetics, so constancy alone does not establish chemical equilibrium.

Consider the idealized elementary chemical reactionAssume a well-mixed closed system at fixed temperature and volume, with first-order law of mass action rates in both directions and no competing chemical reactions. Write $a=[\mathrm{A}]$ and $b=[\mathrm{B}]$ for concentrations in $\mathrm{M}=\mathrm{mol}/\mathrm{L}$. The total $a+b=C=1\mathrm{M}$ is conserved because each conversion replaces one molecule by one molecule. This conservation is special to the chosen stoichiometry.

The forward and reverse reaction rates, measured as conversions per unit volume per unit time, are $J_f=k_{f}a$ and $J_r=k_{r}b$. HenceSetting this derivative to zero givesWith $a(0)=1\mathrm{M}$ and $b(0)=0$, solving the ordinary differential equation yieldsAt $t=5\mathrm{s}$, these are approximately $0.3515\mathrm{M}$ and $0.6485\mathrm{M}$. Initially $J_f=0.3M/s$ and $J_r=0$. At chemical equilibrium, both are $0.075M/s$, while the net reaction rate $J_f-J_r$ is zero. The persistent exchange is visible in the lower panel even after the upper panel has nearly flattened.

Why does chemical equilibrium also involve a free energy minimum? For a closed system at fixed temperature and pressure, without imposed non-expansion work, stable equilibrium minimizes Gibbs free energy over compositions allowed by conserved elemental inventories. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19. This is constrained Gibbs minimization for chemical equilibrium: freely minimizing over unrelated species amounts would change the inventory. At fixed temperature and volume, the appropriate potential is Helmholtz free energy. In either setting, varying an allowed extent of reaction gives the corresponding free energy slope ${\sum }_i{\nu }_i{\mu }_i$, with molar chemical potentials and signed stoichiometric vector entries. An interior equilibrium has zero slope; stability requires a minimum. For A converting to B, this says ${\mu }_{\mathrm{B}}={\mu }_{\mathrm{A}}$.

The beauty of this connection is that molecular exchange, a macroscopic composition, and a geometric minimum describe the same balance. In the ideal concentration model, let $x=b/C$ and choose the standard molar chemical potential difference consistent with $K=3$. Up to a composition-independent constant, the fixed-volume Helmholtz free energy then obeyswhere $n$ is the conserved total amount of substance and $R$ the molar gas constant. Its derivative is $\ln [x/(1-x)]-\ln 3$, vanishing at $x=3/4$. Its second derivative is $1/x+1/(1-x)>0$ inside $0<x<1$, so this balance is the unique minimum. The mixing entropy terms explain why the favored composition still contains A.

The central binomial coefficient and Legendre formula implyfor every sufficiently large real number $x$.

The equation $(1-z^2)w^{\prime \prime }-z{w}^{\prime }+n^{2}w=0$ is obtained from the harmonic equation $w_{xx}+n^{2}w=0$ by $z=\cos x$.

If $f$ has degree at most $n$ and $|f|<1$ on $[-1,1]$, alternation at the extrema of $T_n$ implies $|f(t)|<|T_n(t)|$ outside that interval.

Among monic real polynomials of degree $n$, $2^{1-n}{T}_n$ uniquely minimizes the supremum norm on $[-1,1]$, attaining the value $2^{1-n}$.

Every root of $T_n^{(k)}$ lies in $(-1,1)$ and its leading coefficient is positive. Hencefor $x\ge 1$ and $0\le k\le n$.