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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 $0\le k\le n$.

The number $2\cos (\pi /n)$ is an algebraic integer. If it is rational it must be an integer, which shows that $\cos (\pi /n)$ is rational only for $n=1,2,3$.

The Chebyshev polynomial of the second kind is defined byIt is orthogonal on $[-1,1]$ with weight $\sqrt{1-x^2}$ and has roots $\cos (j\pi /(n+1))$ for $1\le j\le n$.

For a constant uniform magnetic field $B$, choose the symmetric gauge $A=\frac{1}{2}B\times r$. The coordinate parallel to $B$ is cyclic, so its canonical momentum is conserved. Since $A\cdot B=0$, this component is also the mechanical momentum $m\dot{r}\cdot \hat{B}$.

Under $\varphi \mapsto \varphi -{\partial }_{t}f$ and $A\mapsto A+\nabla f$, the charged-particle Lagrangian changes byIt therefore has unchanged equations of motion by total-time-derivative invariance.

Independence makes the characteristic function of a sum equal the product of the individual characteristic functions.

For mutually independent random variables $X_1,\dots ,X_n$, write $S_n={\sum }_{j=1}^n{X}_j$. ThenThis is the expectation of a product of independent random variables applied to bounded complex exponentials; no moment hypothesis is needed.^[ref]

Identical probability distributions are unnecessary. When the random variables are independent and identically distributed random variables, the product is ${\phi }_X(t{)}^n$. A deterministic rescaling also follows directly from the definition: ${\phi }_{aX+b}(t)=e^{itb}{\phi }_X(at)$ for real $a,b$. Consequently the sample mean has characteristic function ${\phi }_X(t/n{)}^n$, and a sum divided by $\sqrt{n}$ has characteristic function ${\phi }_X(t/\sqrt{n}{)}^n$.

The independence assumption does real work. Let $X$ take $-1$ and $1$ with equal probabilities and set $Y=X$. Both characteristic functions are $\cos t$, but ${\phi }_{X+Y}(t)=\cos (2t)$, not ${\cos }^{2}t$. At $t=\pi /2$ they give $-1$ and zero respectively. Knowing the two marginal probability distributions does not determine the sum without their joint probability distribution.

Purely imaginary characteristic roots produce oscillatory solutions. In particular, $r\tau =-\pi /2$ gives $\lambda =\pm i\pi /(2\tau )$ and the real solution $\cos (\pi t/(2\tau ))$.

The degree of an irreducible character $\chi$ is $\chi (1)$, the dimension of the corresponding irreducible representation.

For a discrete memoryless channel with $m$ output symbols, suppose every row of the transition matrix is a permutation of every other row and all column sums agree. Such a communication channel hasThe row Shannon entropy is the same for every input, so $H(Y\mid X)$ has that fixed value for every input probability distribution. The remaining term $H(Y)$ is at most ${\log }_{2}m$ by maximum entropy on a finite alphabet.

A uniform input attains the bound. If there are $\ell$ input symbols, the total of all transition probabilities in the matrix is $\ell$; equal column sums must therefore each be $\ell /m$. Multiplication by the uniform input mass $1/\ell$ gives output mass $1/m$ in each column. Hence $H(Y)={\log }_{2}m$, which proves the displayed formula. The cyclic ternary channel capacity, binary symmetric channel capacity, and ternary symmetric channel capacity are concrete applications of this symmetry calculation.

Center-of-mass reduction separates translational motion from internal coordinates by using the conserved or externally forced center coordinate.

If $G/Z(G)$ is cyclic, then $G$ is abelian. Indeed, if $gZ(G)$ generates the quotient, every element is $g^{a}z$ with $z\in Z(G)$, and any two such elements commute.

The word metric measures how many allowed moves are needed to change one group element into another. For a finite symmetric generating set of a group $S$ with $1\notin S$, it is the graph distance between vertices of the Cayley graph, equivalentlyHere word length counts letters in a shortest representative of $g^{-1}h$. The path metric on the entire Cayley graph also permits points inside edges; the word metric is its restriction to the group itself.^[ref]

The word metric is left invariant:This corresponds to the label-preserving group action on the Cayley graph. Symmetry follows by reversing a shortest graph path; concatenating routes gives the triangle inequality.^[ref]

Right invariance can fail in a non-abelian group, because right multiplication replaces $g^{-1}h$ by $a^{-1}{g}^{-1}ha$ through conjugation. In the free group $F(x,y)$ with $S=\{x,x^{-1},y,y^{-1}\}$, we have $d_S(1,y)=1$, whereas $d_S(x,yx)=|x^{-1}yx{|}_S=3$: the latter freely reduced word has three letters. In an abelian group, conjugation fixes each element, so right invariance does hold. Cayley graphs of the integers and word metric balls in the integer lattice give examples where the word metric can be calculated explicitly.

For an order-$k$ scalar quasilinear partial differential equation with real-analytic coefficients, real-analytic Cauchy data on a real-analytic non-characteristic hypersurface determine a unique real-analytic solution in a neighbourhood of each point of that hypersurface.

For a scalar equation of order $k$, Cauchy data consist of the first $k$ normal jets $u,{\partial }_{N}u,\dots ,{\partial }_N^{k-1}u$ on the initial hypersurface.

These prescribed values define the initial conditions of a Cauchy problem.

For $0<s<a$,