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Expand the relativistic energy as $ϵ(p)=m{c}^2+p^2/(2m)+O(p^4/(m^3{c}^2))$. In the dilute nonrelativistic regime the occupation denominator is dominated by its exponential, so the Fermi-Dirac distribution reduces to the Maxwell-Boltzmann distribution. The number density is thenThe assumed low temperature makes the contributing momenta nonrelativistic; the large positive $(m{c}^2-\mu )/(kT)$ makes occupation numbers small.

For chemical equilibrium of hydrogen ionization, ${\mu }_p+{\mu }_e={\mu }_H+{\mu }_{\gamma }$, and thermal photons have ${\mu }_{\gamma }=0$. Apply the same dilute-gas expression to protons, electrons and ground-state hydrogen. Their density ratio isAssume charge neutrality with no other important ionic species, so $n_p=n_e$; neglect the binding-energy mass correction in $m_p/m_H$, so this ratio is approximately one. Including proton and electron spin gives $g_p=g_e=2$ and ground-state $g_H=4$, hence $g_p{g}_e/g_H=1$. Equivalently one may consistently omit nuclear spin from both proton and hydrogen degeneracies. Thus the Saha equation isThe assumptions are thermal and chemical equilibrium at a common temperature, dilute nonrelativistic species, neutral total charge, predominantly ground-state hydrogen, and negligible interaction/partition-function corrections. The Maxwell-Boltzmann density formula applies to dilute hydrogen whether its composite spin makes it a boson or a fermion.

Noether theorem associates a conserved quantity ${\sum }_i{p}_i\delta q_i$ to a continuous symmetry of the Lagrangian (with the usual boundary-term correction if the Lagrangian changes by a total derivative). The angles $\psi$ and $\varphi$ are cyclic coordinates, so for the heavy symmetric topTime-translation symmetry gives conserved energy, whose Hamiltonian expression isIt is quadratic in the momenta; $p_{\theta }=I_1\dot{\theta }$. Eliminating the cyclic velocities and putting $\mu =\cos \theta$, so ${\dot{\mu }}^2={\sin }^2\theta {\dot{\theta }}^2$, yieldsThe first term has degree three and the second degree two, displaying the cubic governing nutation.

The numbers $z_j$ are the nonzero zeros of a function $f$, repeated according to their multiplicity. The displayed Weierstrass factorization theorem formula uses genus-one Weierstrass elementary factors and a linear exponential factor; it applies under the corresponding growth assumptions, rather than to every arbitrary entire function.

Apply it to the entire function $g(z)=\sin (\pi z)/(\pi z)$, with its removable value $g(0)=1$. Its zeros are $\pm 1,\pm 2,\dots$, each simple, and $g^{\prime }(0)=0$ because $g$ is even. Pairing the factors at $n$ and $-n$ givessoThe product converges uniformly on compact subsets because $\sum n^{-2}<\infty$. Away from the integers its logarithmic derivative can be taken term by term: the resulting series also converges locally uniformly, its terms being $O(n^{-2})$. HenceThis is an identity of meromorphic functions, understood away from their poles.

The physical state space is the positively invariant set $X,Y,Z\ge 0$, $X+Y+Z=N$: adding the equations gives zero, and the vector field points inward at each coordinate boundary. At an equilibrium point, either $Y=0$, giving $(X,Y,Z)=(N,0,0)$, or $\beta X=b+r$. ThereforeA positive endemic equilibrium exists exactly when $N>N_c$.

For $N<N_c$, put $d=b+r-\beta N>0$. Since $X\le N$,The last limit follows either by evaluating the exponential integral (with a $t{e}^{-bt}$ term when $d=b$), or by splitting it into an early decaying part and a late part with uniformly small $Y$. Hence $X\to N$ for every physical initial condition. Although immune numbers need not decrease monotonically initially, both infected and immune numbers tend to zero.

At the endemic equilibrium, the Jacobian matrix of the $X,Y$ subsystem isA disturbance proportional to $e^{pt}$ therefore obeysBoth coefficients are positive. Real roots have negative sum and positive product; complex conjugate roots have real part $-(b+\beta Y_s)/2<0$. Thus the endemic equilibrium is asymptotically stable by linear stability analysis.

A generalized linear model has independent responses whose distributions belong to an exponential family:Here the expected value and variance are ${\mu }_i=b^{\prime }({\theta }_i)$ and $\mathrm{Var}(Y_i)=a_i(\varphi )b^{\prime \prime }({\theta }_i)$. A specified one-to-one link function $g$ connects the mean to the linear predictor: $g({\mu }_i)=x_i^T\beta$. The canonical link function is $g({\mu }_i)={\theta }_i$.

In binomial regression, $Y_i\sim \mathrm{Bin}(m_i,p_i)$ independently, with known trial counts $m_i$ and probabilities related to the predictor. The logistic regression and probit regression specifications are respectivelyHere $\Phi$ is the standard normal distribution function. The binomial log-density contains $y_i\log (p_i/(1-p_i))+m_i\log (1-p_i)$, so its natural parameter is the log-odds: the logistic link is canonical. If the response mean is written as ${\mu }_i=m_i{p}_i$, the canonical link is $g_i({\mu }_i)=\log ({\mu }_i/(m_i-{\mu }_i))$.

For a binary linear code $C\le {\mathbb{F}}_2^n$, its rank is its dimension of a vector space $r$. Use the weight enumerator conventionwhere $w(c)$ is the Hamming weight. This convention is essential here: $W_C(1,0)$ counts words of weight $n$, rather than zero words. There are $2^r$ codewords, so $\boxed{W_C(1,1)=2^r}$.

There is only one binary word of weight $n$, namely $1=(1,\dots ,1)$. Thus $W_C(1,0)=1$ precisely when $1\in C$. If this holds, the linear map $c\mapsto c+1$ is a bijection of $C$ and replaces weight $w$ by $n-w$; hence $W_C(s,t)=W_C(t,s)$. Conversely this symmetry gives $W_C(1,0)=W_C(0,1)=1$, because the zero word is the unique word of weight zero.

The repetition code consists of $0,1$, givingThe single parity-check code consists of all words of even weight. Exactly $\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ words have weight $j$, so the binomial theorem gives

A Kleinian group is a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$, acting by Möbius transformations on the Riemann sphere, equivalently by orientation-preserving isometries of hyperbolic space ${\mathbb{H}}^3$.

Consider the classes ofThey generate a discrete subgroup because their matrix representatives have integer entries: a sequence of such matrices converging to $I$ or $-I$ is eventually constant. On the real projective line set $X=\{x:|x|>1\}\cup \{\infty \}$ and $Y=\{x:|x|<1\}$. For every nonzero integer $m$,The last inclusion follows from $|2mx+1|\ge 2|x|-1>|x|$ when $|x|>1$; also $B^m(\infty )=1/(2m)\in Y$. The ping-pong lemma therefore says that the subgroup is the free product $⟨A⟩\ast ⟨B⟩$. Both generators have infinite order, soa free group on two generators. The ping-pong inclusions ensure that no nonempty reduced word can act as the identity, which is the essential freeness argument.

A plane wave in the rest frame of the medium has phase $\mathbf{k}\cdot \mathbf{x}-{\Omega }_0(\mathbf{k})t$. Under the Galilean transformation ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{U}t$, this becomes $\mathbf{k}\cdot {\mathbf{x}}^{\prime }-[{\Omega }_0(\mathbf{k})-\mathbf{U}\cdot \mathbf{k}]t$. ThereforeHere ${\mathbf{c}}_g$ is the group velocity in the moving frame.

For acoustic waves, ${\Omega }_0=c_0|\mathbf{k}|$ and take $\mathbf{U}=M{c}_0{\mathbf{e}}_x$. A steady wave has $\omega =0$, requiring $\cos \beta =1/M$, where $\beta$ is the angle of $\mathbf{k}$ to the flight direction. Its ray velocity is $c_0\hat{\mathbf{k}}-M{c}_0{\mathbf{e}}_x$, with backward component $-c_0(M^2-1)/M$ and transverse magnitude $c_0\sqrt{M^2-1}/M$. Thus the angle $\theta$ of the ray to the backward flight direction obeysThe axial symmetry gives the Mach cone behind the aircraft.

For unsteady emitted disturbances, all ray velocities lie on the sphere ${\mathbf{c}}_g=-\mathbf{U}+c_0\hat{\mathbf{k}}$. Since $M>1$, the origin lies outside this velocity sphere and its tangent cone has exactly the same semi-angle. All other rays are strictly inside the cone. Equivalently a signal emitted $\tau$ time units ago has a wavefront sphere centered at $-\mathbf{U}\tau$ with radius $c_0\tau$; these spheres are enclosed by their Mach-cone envelope. Consequently the unsteady wavefield is confined to the cone's interior, with the steady envelope on its boundary.

For capillary waves put $a=\sqrt{T/\rho }$, so ${\Omega }_0=a{k}^{3/2}$. The positive-frequency steady condition is $a\sqrt{k}=U\cos \beta$, requiring $\cos \beta >0$. ThenandIts endpoint traces the circlesince $c_{gx}/U=-1/4+(3/4)\cos 2\beta$ and $c_{gy}/U=(3/4)\sin 2\beta$. As $\beta$ runs through $(-\pi /2,\pi /2)$ this traces the circle except its leftmost limiting point, corresponding to the zero-wavenumber limit. The circle surrounds the origin, so every direction except the precisely backward axis is attained by a nonzero-wavenumber steady ray; the backward direction is approached as $k\downarrow 0$. Thus the closure of the steady ray field extends all around the insect, with no excluded angular sector such as the acoustic Mach-cone exterior.

Write $D={\partial }_n{A}_n$ and use repeated-index summation. Because every $A_i$ is a harmonic function,andThus $\mu {\nabla }^2{u}_i-{\partial }_{i}p=0$, the incompressible Stokes equation. MoreoverThe antisymmetric first two terms cancel in the sum with ${\partial }_i{u}_j$, giving the Newtonian fluid Cauchy stress tensorFor $A_i=V_i(1-a/(2r))$, harmonic outside the origin,At $r=a$ this gives $\mathbf{u}=\frac{1}{2}[\mathbf{V}-(\mathbf{V}\cdot \mathbf{n})\mathbf{n}]$, where $\mathbf{n}=\mathbf{x}/a$, so the normal velocity is zero. AlsoHence on the surface,a purely normal traction, appropriate to a shear-free sphere rather than a no-slip solid sphere. Finally rotational symmetry gives ${\int }_{S^2}{n}_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$, so the fluid's force on the sphere isThis is the clean-bubble Stokes drag in the far-field velocity $\mathbf{V}$.

Use SI units, $x^a=(ct,\mathbf{x})$ and metric signature $(+,-,-,-)$. The four-current is $j^a=(c\rho ,\mathbf{j})$. Charge density and current transform together under Lorentz transformations, and the continuity equation becomesWith the four-potential $A^a=(\phi /c,\mathbf{A})$, the Lorenz gauge ${\partial }_a{A}^a=0$ converts Maxwell equations to $\square A^a={\mu }_0{j}^a$, where $\square =c^{-2}{\partial }_t^2-{\nabla }^2$. Choosing no incoming homogeneous radiation, the retarded electromagnetic potential isCurrent conservation ensures this solution obeys the Lorenz gauge; a homogeneous solution may be added when additional field boundary data are specified.

For a thin loop, let $R=|\mathbf{x}|$, $\hat{\mathbf{R}}=\mathbf{x}/R$ and $t_R=t-R/c$. ThenIn the small-loop magnetic-dipole approximation, expand $|\mathbf{x}-{\mathbf{x}}^{\prime }|=R-\hat{\mathbf{R}}\cdot {\mathbf{x}}^{\prime }+\cdots$. The constant term integrates to zero because $\oint d{\mathbf{\ell }}^{\prime }=0$. The first nonzero term isFor the oriented circular loop the last integral is $\pi r^2\mathbf{n}\times \hat{\mathbf{R}}$. Thus with magnetic dipole moment $\mathbf{m}(t)=\pi r^{2}I(t)\mathbf{n}$,The $R^{-2}$ term is the near dipole potential and the $R^{-1}$ term its radiation potential. This is the leading small-$r$ result at fixed $R,\omega$; besides $r/R\ll 1$, the expansion of the retarded current requires $\omega r/c\ll 1$.

If only $r/R\ll 1$ is imposed and the loop need not be small compared with a wavelength, keep the retarded phase around the loop. For fixed $r,\omega$ and $R\to \infty$, the leading far-field expression is insteadwhere $\cos \theta =\mathbf{n}\cdot \hat{\mathbf{R}}$ and $J_1$ is the Bessel function of the first kind. Integrating the sinusoidal phase on a circle gives this Bessel factor; $J_1(z)\sim z/2$ recovers the radiation part of the boxed dipole result.

Assume $A>0$ and $\alpha >0$, so the excitation energy increases with momentum and vanishes in the ground state. With one internal state, the number of states in a spherical momentum shell is $V4\pi k^{2}dk/(2\pi {)}^3$. Since $k=(ϵ/A{)}^{1/\alpha }/\hslash$,An internal degeneracy multiplies this density of states by its degeneracy factor.

The Bose-Einstein distribution is $\bar{n}(ϵ)=[e^{(ϵ-\mu )/(k_{B}T)}-1{]}^{-1}$. For a finite nonnegative occupation of the zero-energy state, its denominator must be positive, requiring $\mu <0$. At Bose-Einstein condensation the thermodynamic-limit chemical potential approaches $0$ from below, and the ground state must be counted separately; therefore “$\mu <0$” is a finite-system condition, while the condensed limit has $\mu =0$.

In a box whose shape is fixed, each momentum quantum number scales as $V^{-1/3}$, so each energy level scales as $V^{-\alpha /3}$. Differentiating the grand potential $\Omega =k_{B}T{\sum }_j\log (1-e^{-({ϵ}_j-\mu )/(k_{B}T)})$ at fixed $T,\mu$ givesThe zero-energy condensate contributes neither pressure nor kinetic energy.

The number outside the ground state isAt $\mu \uparrow 0$, its low-energy integrand is proportional to ${ϵ}^{3/\alpha -2}$. The integral is finite precisely when $3/\alpha >1$, soFor these exponents, the maximum excited population isThis follows by expanding the Bose denominator in positive exponentials and integrating each term. It tends to zero as $T\downarrow 0$, forcing excess particles into the ground state at fixed nonzero total number. At $\alpha =3$ the divergence is logarithmic; for $\alpha >3$ it is stronger, so there is no positive-temperature thermodynamic-limit condensation of this homogeneous gas.

Photons have $\alpha =1$ and two polarizations, but in ordinary blackbody equilibrium their number is not conserved and $\mu =0$ at every temperature. Cooling reduces their number rather than forcing an excess into a ground state, so an ordinary equilibrium photon gas does not condense. Systems imposing an effective photon-number constraint and an appropriate confined spectrum are different from the unconstrained photon gas considered here.

Bloch theorem states that energy eigenfunctions in a Bravais lattice potential may be chosen asThe wavevector is defined modulo a reciprocal lattice vector, so one can restrict it to the first Brillouin zone. This follows because lattice translations commute with the Hamiltonian and have simultaneous eigenvalues $e^{i\mathbf{k}\cdot \mathbf{l}}$.

In one dimension write $V(x)={\sum }_G{V}_G{e}^{iGx}$ with $G=2\pi j/a$ and expand a Bloch state as ${\psi }_k={\sum }_G{c}_G{e}^{i(k+G)x}$. The Schrödinger equation becomesFor zero potential, the parabolas $E_G^0(k)={\hslash }^2(k+G{)}^2/(2m)$ fold into the Brillouin zone. A weak potential shifts a nondegenerate level by $V_0$ at first order, with second-order correction ${\sum }_{G\ne 0}|V_G{|}^2/(E_0^0-E_G^0)$ where the denominators stay away from zero. This expansion fails near crossings, where degenerate perturbation theory is required.

At the first zone boundary, put $G=2\pi /a$ and $k=G/2+q$. The nearly degenerate waves with momenta $\hslash k$ and $\hslash (k-G)$ are coupled by $V_G$. Retaining their two-dimensional subspace givesDiagonalizing producesAt $q=0$ there is a gap $\boxed{\Delta E=2|V_G|}$: the two standing-wave combinations sample the periodic potential differently. Similar Bragg degeneracies occur at $k=j\pi /a$, with gaps controlled to first order by the corresponding Fourier components. If that component vanishes, a gap need not open at first order. Between the gaps the continuous energy eigenvalues form energy bands, periodically represented in the reduced-zone scheme. For the first zone edge, the curvature of the upper and lower branches gives a minimum and a maximum respectively for a sufficiently weak nonzero potential, so the avoided crossing is a genuine local forbidden energy interval.

At zero temperature electrons occupy states up to the Fermi energy. The Fermi surface is the set $E_n(\mathbf{k})=E_F$ separating occupied from unoccupied states within partially filled bands (Fermi points in one dimension). A partially filled band supplies arbitrarily nearby empty states, allowing occupations to respond to a weak electric field and conduct. Completely filled bands give no net current from such a redistribution; if the next empty band is separated by a positive band gap, the system is a band insulator. Band overlap can instead yield a metal even at an integer band filling. At finite temperature an insulator can conduct through thermally excited electrons and holes.