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Consider two internally equilibrated reservoirs A and B at the same imposed temperature and pressure. They may have different compositions or represent different phases. Transfer a small positive amount of substance $\delta n$ of the same chemical species from A to B, holding the other inventories fixed. To first order, their total Gibbs free energy changes byHence exchange toward lower chemical potential lowers Gibbs free energy. If transfer is allowed in both directions and the two reservoirs contain the exchangeable component, thermodynamic equilibrium requires equal chemical potentials for that component. See MIT 5.60, Lecture 14, pages 1–2.

An impermeable partition can maintain unequal chemical potentials; equality is a condition on an allowed exchange. Two pure samples of the same ideal gas already have equal chemical potentials at the same temperature and pressure. Different compositions at a common pressure can instead give different chemical potentials for one component through its thermodynamic activity.

Equality concerns the same exchangeable component in different places, not all different chemical species. For a permitted chemical reaction, the corresponding interior chemical equilibrium criterion is ${\sum }_i{\nu }_i{\mu }_{\mathrm{m},i}=0$, with signed stoichiometric vector entries ${\nu }_i$. The existing chemical equilibrium and constrained Gibbs minimization for chemical equilibrium sections develop those composition constraints.

A family of $n+1$ real continuous functions is a Chebyshev system exactly when every evaluation matrix at $n+1$ distinct points is invertible. A singular evaluation matrix supplies a nonzero coefficient vector whose function vanishes at those points; conversely, a violation of the zero bound supplies such a vector in the kernel. This also gives the existence and uniqueness of interpolation from its span.

Exponentials with distinct real exponents form a Chebyshev system on every nondegenerate real interval. For an inductive proof, multiply a candidate combination by $e^{-{\lambda }_{0}x}$. If it has $n+1$ zeros, Rolle's theorem gives at least $n$ zeros of its derivative, a combination of the $n$ exponentials with exponents ${\lambda }_i-{\lambda }_0$, $i>0$. The inductive zero bound forces all those coefficients to vanish. The remaining constant also vanishes, a contradiction.

A Chebyshev system of real continuous functions on a circle has odd dimension. If its dimension is $d$, take $d$ cyclically ordered evaluation points and move them continuously to the next cyclic permutation without letting points collide. The evaluation determinant stays nonzero, so cannot change sign. The final cyclic row permutation multiplies it by $(-1{)}^{d-1}$; consequently $d$ must be odd. For example, $1,\cos x,\sin x,\dots ,\cos mx,\sin mx$ has dimension $2m+1$ and the required zero bound.

Let $x_j=\cos ((2j-1)\pi /(2n))$ be the roots of the Chebyshev polynomial, and let $q$ have degree at most $n-1$ with $|q(x_j)|\le |T_n^{\prime }(x_j)|$. Its Lagrange interpolation polynomial uses basis ${\ell }_j(x)=T_n(x)/[(x-x_j)T_n^{\prime }(x_j)]$. For $x>x_1$, $T_n(x)>0$ and $x-x_j>0$, so ${\ell }_j(x)T_n^{\prime }(x_j)>0$. Therefore $|q(x)|\le {\sum }_j|{\ell }_j(x)T_n^{\prime }(x_j)|=T_n^{\prime }(x)$, the last equality being interpolation of the derivative. Continuity includes $x=x_1$ and reflection proves the left-hand region.

For weighted least-squares polynomial in an orthogonal-polynomial basis with weight $(1-x^2{)}^{-1/2}$ on $[-1,1]$, the projection of $\sqrt{1-x^2}$ onto polynomials of degree at most $n$ is the displayed sum. Put $x=\cos \theta$ to obtain $T_j(x)=\cos j\theta$ and $f(x)=\sin \theta$ on $[0,\pi ]$. The constant coefficient is $2/\pi$, odd coefficients vanish, and the coefficient of $T_{2m}$ is $4/[\pi (1-4m^2)]$. For odd $n$ the optimal polynomial can have degree $n-1$.

The lowest Lambda c baryon has valence content $cud$, electric charge $+1$, isospin zero and spin angular momentum $1/2$. Its light pair has antisymmetric flavour and spin zero. This is the charm analogue of the Lambda baryon.

The symmetric light pairs $uu,\{ud\},dd$ with one charm quark give isospin one and charges $2,1,0$. Their light spin is one; coupling the charm spin gives the $J=1/2$ Sigma c baryon triplet and $J=3/2$ excited-spin triplet conventionally marked by a star.

Linear charge density is electric charge divided by length along a charged line, or locally $dQ/d\ell$. Its SI unit is coulombs per metre. For a uniformly charged long cylindrical conductor, it equals the total surface charge density integrated around the cylinder's circumference; it need not represent a charge literally supported on a one-dimensional line.

Charge conjugation symmetry is invariance of the dynamics under charge conjugation. It pairs a process with its particle-antiparticle conjugate at equal rates. C violation is the failure of this invariance.

Average $1-\cos (uy)$ over $0\le u\le 1/\lambda$. Multiplication by $\lambda$ gives $1-\sin (y/\lambda )/(y/\lambda )$, with value zero at $y=0$. This is at least $1-\sin 1$ when $|y|\ge \lambda$. Tonelli's theorem then gives the displayed bound for a probability measure with characteristic function $\varphi$. The bound converts control of the characteristic function near zero into tightness of probability measures and is useful in proving the Lévy continuity theorem.

If $\mathbb{E}|X|<\infty$, the characteristic function $\varphi (u)=\mathbb{E}{e}^{iuX}$ satisfies ${\varphi }^{\prime }(0)=i\mathbb{E}X$. The bound $|(e^{iuX}-1)/u|\le |X|$ and the dominated convergence theorem justify differentiation at zero. For mean zero, $\psi (u)=|1-\varphi (u)|/|u|$ extends continuously by $\psi (0)=0$, and $|1-\varphi (u{)}^n|\le n|u|\psi (u)$ by a finite geometric series. This supplies a direct characteristic-function proof of the weak law of large numbers.

// Target: martingale.bigb

For a constant-coefficient linear recurrence relation over the complex numbers, a nonzero root $\mu$ of multiplicity $m$ in its characteristic polynomial of a linear recurrence supplies the $m$ linearly independent solutionsSubstitute $y_n={\mu }^n{z}_n$. Each factor $E-\mu$ of the shift linear operator becomes $\mu$ times the forward difference operator on $z_n$, so its $m$th power kills every polynomial in $n$ of degree less than $m$. Their independence follows from the Lagrange root bound over a field. The same argument works over a field of characteristic zero; the displayed monomials need modification in positive characteristic.

A zero root needs separate initial-data treatment: for characteristic polynomial ${\lambda }^m$, the recurrence relation is $y_{n+m}=0$, leaving the first $m$ values arbitrary.

Every complex character of a representation of a finite group $G$ is an integer linear combination of characters of induced representations from one-dimensional group representations of subgroups $H_j\le G$. Negative coefficients are allowed. Unlike Artin induction, the subgroups need not be cyclic groups. Applied to an Artin L-function, compatibility with induced representations turns this character identity into a product of one-dimensional Artin L-functions with integer exponents, reducing the behavior at $s=1$ to abelian extensions.

Every complex character of a representation of a finite group $G$ is a rational linear combination of characters of induced representations from one-dimensional group representations of cyclic subgroups $C_j$. One-dimensionality follows because the irreducible representations of a finite cyclic group over the complex numbers are one-dimensional. The rational coefficients distinguish this statement from Brauer induction, which permits integer coefficients by allowing more general subgroups. For a rational-valued character, another form of Artin induction uses rational combinations of induced representations of trivial group representations of cyclic subgroups.

A group of automorphisms of a finite group permutes both its conjugacy classes and its irreducible characters. The two permutation representations have the same character: each automorphism fixes equally many conjugacy classes and irreducible characters. The invertible character table intertwines the two permutation matrices, which have equal traces. The claim concerns fixed classes, not merely fixed elements.

Column orthogonality and the regular character give quadratic and linear equations for missing character values once their degrees are known. These equations can admit spurious rational solutions. Every character value is an algebraic integer, so a rational candidate must be an ordinary integer. Applying this extra condition can remove ambiguities that orthogonality alone leaves.

The irreducible characters of a finite group form a basis of its complex class functions. Equality of every irreducible character value therefore forces equality of every class-function value, including indicators of conjugacy classes, and hence conjugacy. The converse follows from the class-function property.

The pulsation period of a Cepheid variable is correlated with its absolute magnitude in a specified photometric band. Calibrating the relation with stars of known distance allows an observed period and corrected apparent magnitude to give a distance. Extinction and metallicity corrections matter for a precise cosmic distance ladder.

Put $c=m\kappa /{\hslash }^2$, $k_n=(n+1/2)\pi /a$, and ${\delta }_n=a({\lambda }_n-k_n)$. The quantization equation gives ${\delta }_n=\arctan (c/{\lambda }_n)$, so ${\lambda }_n{\delta }_n\to c$ and ${\delta }_n\to 0$. Therefore ${\lambda }_n^2-k_n^2\to 2c/a$, proving the displayed absolute energy shift. Its ratio to the unperturbed energy tends to zero. The normalized unperturbed even state has probability density $1/a$ at the origin, so its first-order delta-barrier expectation is also $\kappa /a$.