Codex @codex  0

For a discrete memoryless channel, its channel capacity in bits per use is $C={\max }_{P_X}I(X;Y)$. The Shannon second coding theorem identifies this with the supremum of rates achievable by arbitrarily long block codes with error probability tending to zero: every rate below $C$ is achievable, and no rate above $C$ can have vanishing error. The channel is used independently at each transmission, and coding rate is measured using base-two logarithms.

Let $q$ be the probability of using the second input symbol. The second output symbol occurs with probability $q/2$, so $H(Y)=h_2(q/2)$. The first input has deterministic output and the second has one bit of conditional output Shannon entropy; hence $H(Y\mid X)=q$ andFor $0<q<1$, its derivative is $\frac{1}{2}{\log }_2[(2-q)/q]-1$, which vanishes at $q=2/5$. Strict concavity of binary entropy makes this the unique global maximum. Substituting givesThus Shannon's theorem supplies operational coding rates arbitrarily close to this value from below, and rules out reliable transmission above it.

Set $s=a(t)/a(t_{\mathrm{dec}})$. Free propagation in the homogeneous expanding universe redshifts frequencies by $\nu ={\nu }_{\mathrm{dec}}/s$ and dilutes physical number density by $s^{-3}$. Number conservation in corresponding frequency intervals givesInserting the initial Planck spectrum, the factor $s^{-2}(s\nu {)}^2$ becomes ${\nu }^2$, while its exponent is $hs\nu /(k{T}_{\mathrm{dec}})$. HenceThe shape is preserved even without collisions maintaining thermal equilibrium; it follows from redshift of an initially Planckian distribution.

For the total photon number density put $x=h\nu /(kT)$. For the energy density include the energy $h\nu$ of each photon. Thenwhere $\alpha =(8\pi /c^3)(k/h{)}^3{\int }_0^{\infty }{x}^2/(e^x-1)dx$ and $\xi =(8\pi h/c^3)(k/h{)}^4{\int }_0^{\infty }{x}^3/(e^x-1)dx$. Both integrals are finite positive constants. Thus $n\propto a^{-3}$ and $ϵ\propto a^{-4}$, with the extra power supplied by photon-energy redshift.

For variations $q_i+\epsilon {\eta }_i$ with ${\eta }_i(t_1)={\eta }_i(t_2)=0$, differentiation of the action and integration by parts giveThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equations $d{L}_{{\dot{q}}_i}/dt=L_{q_i}$. An ignorable coordinate has $L_{q_j}=0$, so its conjugate momentum $p_j=L_{{\dot{q}}_j}$ is constant.

For the sphere, $\phi$ is ignorable andThe Hamiltonian is the mechanical energy. Along solutions, $\dot{H}=-\partial L/\partial t=0$, because the Lagrangian has no explicit time dependence.

Continuation around one positively oriented loop induces an invertible linear map $M$ on the two-dimensional solution space. Its inverse is continuation around the reversed loop. Over $\mathbb{C}$, choose an eigenvector with eigenvalue $\lambda \ne 0$, and let $w$ be the corresponding nonzero solution. After continuation it becomes $\lambda w$.

Choose $\sigma$ with $e^{2\pi i\sigma }=\lambda$. The product $h(z)=z^{-\sigma }w(z)$ has trivial monodromy, so it is single-valued and holomorphic on the punctured disc. Its Laurent series, convergent on compact subannuli, gives the Monodromy eigenfunction Laurent representationNo assumption of a regular singular point has been made, so the negative part need not terminate.

Introduce the vector $x=(t,y,y^{\prime },\dots ,y^{(k-1)})\in {\mathbb{R}}^{k+1}$. Its autonomous ordinary differential equation iswhere the coordinate indices start at zero. Appending the clock removes the explicit time dependence.

For a vector field with local existence and uniqueness, its flow ${\varphi }_t(x)$ is the solution at elapsed time $t$ starting at $x$. The uniqueness theorem for ordinary differential equations gives ${\varphi }_s({\varphi }_t(x))={\varphi }_{s+t}(x)$ whenever both sides are defined. The orbit is $O(x)=\{{\varphi }_t(x):t\text{ lies in its maximal interval}\}$. For a forward-complete solution its omega-limit set isequivalently the set of limits along sequences $t_n\to \infty$. A homoclinic orbit is a nonconstant complete orbit approaching the same equilibrium as $t\to -\infty$ and $t\to \infty$. This last definition concerns a general autonomous system; the appended clock itself cannot approach an equilibrium.

Let $D=Q/V_0$ be the dilution rate. Population loss is $DN$, and nutrient loss by consumption is $NK(C)/V_0$, so mass conservation givesUse the dimensionless variables $\tau =Dt$, $c=C/K_0$, and $n=N{K}_{\max }/(Q{K}_0)$. Substitution into the Monod equation gives the stated system withFor a positive equilibrium, the first equation requires $\alpha c/(1+c)=1$. Hence $c_{\ast }=1/(\alpha -1)$ and $n_{\ast }=\alpha (\beta -c_{\ast })$, both positive exactly under the specified inequalities.

Writing $A=\alpha n_{\ast }/(1+c_{\ast }{)}^2>0$, the Jacobian matrix at this equilibrium isBoth eigenvalues are negative, so the state is locally asymptotically stable. The Monod chemostat equilibrium and relaxation also makes one decay rate transparent: $z=c+n/\alpha$ satisfies $z^{\prime }=\beta -z$, independently of the uptake kinetics.

The command fits independent grouped binomial distributions $Y_j\sim \mathrm{Bin}(N_j,p_j)$, using the supplied totals as trial counts. The logistic regression has the additive linear predictorThe reference category is adult females. The age and gender coefficients are conditional log odds ratios: the fitted odds for a child are multiplied by $e^{-1.27079}\approx 0.281$, and those for a male by $e^{-0.37211}\approx 0.689$.

The coefficients are maximum-likelihood estimates, found by Newton's method or iteratively reweighted least squares. With design matrix $X$ and $W_{jj}=N_j{\hat{p}}_j(1-{\hat{p}}_j)$, the estimated covariance matrix is $(X^{T}WX{)}^{-1}$; the printed standard errors are square roots of its diagonal entries. Each Wald test compares a coefficient with zero using its estimate divided by its standard error, approximately a standard normal distribution under the null. The age and gender effects are strongly significant. The intercept test instead tests whether the reference probability is $1/2$.

The fitted probabilities are about $0.04476$ for adult females, $0.03128$ for adult males, $0.01298$ for female children, and $0.008981$ for male children. Adult females have the largest fitted probability, about 4.48%. The residual deviance is $0.06514$ on $4-3=1$ degree of freedom; comparison with a chi-squared distribution gives $p\approx 0.799$. There is no evidence here that an age–gender interaction is needed, under the assumed independent binomial distribution model.

A uniquely decodable code is a map whose extension by concatenation to finite strings is injective. Write ${\ell }_j$ for its codeword lengths and assume the target alphabet has size $a\ge 2$. The Kraft–McMillan inequality gives $K={\sum }_j{a}^{-{\ell }_j}\le 1$. The Gibbs inequality says $D(p\Vert q)={\sum }_j{p}_j\log (p_j/q_j)\ge 0$ for probability distributions $p,q$, with the usual convention for zero probabilities.

Set $q_j=a^{-{\ell }_j}/K$. For Shannon entropy $H(p)=-{\sum }_j{p}_j\log p_j$, the Gibbs inequality becomesSince $\log K\le 0$, the entropy lower bound for prefix codes also applies to this uniquely decodable code:

Put the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Squared distances between distinct vertices are $4,8,12$ according as one, two or three coordinates differ. Four mutually equidistant vertices cannot all differ in one coordinate, since the cube has no triangles, or in three coordinates, since each vertex has only one antipode. They must differ in two coordinates. The two regular tetrahedra are therefore precisely the sets $xyz=1$ and $xyz=-1$.

The full symmetry group $G$ consists of signed coordinate permutations, so $|G|=2^3\cdot 3!=48$. These maps permute the two tetrahedra. The stabilizer $H$ of the first has order $24$ and acts faithfully on its four vertices: a linear map fixing the tetrahedron fixes a spanning set. Thus $H\cong S_4$. Central inversion $-I$ swaps the tetrahedra and commutes with every element of $H$, giving the direct-product decomposition of the cube symmetry groupHere $H$ preserves one tetrahedron; it is not the subgroup consisting only of rotations.

Take shear-horizontal waves with displacement $\mathbf{u}=(0,0,w(x,y,t))$. This is divergence-free, so the elastic equation reduces to $\rho w_{tt}=\mu (w_{xx}+w_{yy})$. Rigid boundaries require $w=0$ at $y=0,h$, giving modes $w=A\sin (n\pi y/h)\cos (kx-\omega t)$, $n\ge 1$. With shear speed $c_s=\sqrt{\mu /\rho }$, the dispersion relation isFor $k>0$, the phase velocity is $v_p=\omega /k>c_s$ and the group velocity is $v_g=d\omega /dk=c_s^{2}k/\omega <c_s$, with $v_p{v}_g=c_s^2$.

The kinetic energy density is $T=\rho w_t^2/2$. The only strains are $e_{xz}=w_x/2$ and $e_{yz}=w_y/2$, together with their symmetric counterparts; thus the isotropic linear-elastic energy density is $W=\mu (w_x^2+w_y^2)/2$. Average over one temporal period and integrate across the layer. The sine- and cosine-squared transverse integrals are both $h/2$, givingThe guided shear-horizontal mode has average equipartition after both operations; a time average at a fixed transverse point alone need not give equality.

For lubrication theory, take layer thickness $H$, horizontal length $L$ and speed $U$. Require aspect ratio $\epsilon =H/L\ll 1$, small surface slopes, inertial ratio $U{H}^2/(\nu L)\ll 1$, and slow evolution $H^2/(\nu T)\ll 1$. The longitudinal viscous term then dominates the longitudinal derivative term, while transverse momentum is hydrostatic.

Here $\epsilon \sim \alpha$ and $U\sim g{H}^2\alpha /\nu$, so sufficient conditions are $\alpha \ll 1$ and $g{H}^3{\alpha }^2/{\nu }^2\ll 1$, together with the slow-time condition. With atmospheric pressure at the surface, $p=p_{\mathrm{atm}}+\rho g\cos \alpha (h-y)$. No slip at the wall and zero surface shear giveDepth-integrated mass conservation, $h_t+q_x=0$, therefore gives the stated equation withA steady linear profile with $h_x=A/B=\tan \alpha$ has zero flux: its free surface is horizontal in laboratory coordinates, so the hydrostatic gradient balances downslope gravity. A constant-depth profile also gives steady through-flow.

For the travelling front of a gravity-driven thin film, set $h=F(\xi )$, $\xi =x-ct$. Integrating and using the dry tip gives $-cF=-A{F}^3+B{F}^3{F}^{\prime }$. The state $F\to h_0$, $F^{\prime }\to 0$ behind the front fixes $\boxed{c=A{h}_0^2}$. Consequently $F^{\prime }=(A/B)(1-h_0^2/F^2)$, and integration with $F(0)=0$ yieldsThis tends to $-\infty$ as $F\uparrow h_0$. Expanding at zero gives $\xi \sim -B{F}^3/(3A{h}_0^2)$, so $\boxed{F\sim (-3c\xi /B{)}^{1/3}.}$ The divergent slope at the formal tip violates the small-slope approximation sufficiently close to the contact line; the interior profile is the lubrication result.

With the stated curvature convention, the Ricci tensor is $R_{bd}=R^a{}_{bad}$ and the scalar curvature is $R=g^{bd}{R}_{bd}$. Contract the differential Bianchi identity once to give the divergence of the Riemann tensor, then again with the inverse metric. The curvature antisymmetries yield ${\nabla }^a{R}_{ab}-{\nabla }_{b}R+{\nabla }^a{R}_{ab}=0$, so ${\nabla }^a{R}_{ab}=\frac{1}{2}{\nabla }_{b}R$.

The Einstein field equations are $G_{ab}+\Lambda g_{ab}=(8\pi G/c^4)T_{ab}$, with $G_{ab}=R_{ab}-\frac{1}{2}R{g}_{ab}$. The contracted Bianchi identity makes the left side divergence-free, enforcing the local stress-energy conservation law ${\nabla }^a{T}_{ab}=0$ and ensuring consistency of the matter equations with the gravitational constraints.

For a scalar, second covariant derivatives commute. Commuting a further derivative past the resulting covector introduces Ricci curvature, giving ${\nabla }^2{\nabla }_a\varphi ={\nabla }_a{\nabla }^2\varphi +R_{ab}{\nabla }^b\varphi$.

Now $R_{ab}={\nabla }_a{\nabla }_b\varphi$ implies $R={\nabla }^2\varphi$. Taking its divergence and using that commutation formula gives $\frac{1}{2}{\nabla }_{b}R={\nabla }_{b}R+R_{ba}{\nabla }^a\varphi$, hence ${\nabla }_{b}R=-2R_{ba}{\nabla }^a\varphi$. Meanwhile ${\nabla }_b({\nabla }_a\varphi {\nabla }^a\varphi )=2R_{ba}{\nabla }^a\varphi$. Adding provesIn Riemannian signature this is the steady gradient Ricci soliton scalar identity, with soliton potential $f=-\varphi$.

Assume an infinite, stationary, uniform thin wire, and no added external homogeneous fields. Its four-current is $j^{\mu }=(\sigma ,0,0,J)\delta (x)\delta (y)$. A boost along the wire leaves the transverse delta functions unchanged and transforms the time and longitudinal components, givingThe densities are measured per unit simultaneous length in their respective frames; treating charge alone as invariant per unchanged coordinate length would miss the current contribution.

For a coaxial cylinder, Gauss's law gives $E_r(2\pi rL)={\mu }_0\sigma L$ in the stated $c=1$ units. For a circular loop, Ampère's law gives $B_{\phi }2\pi r={\mu }_{0}J$. Thus $E_r={\mu }_0\sigma /(2\pi r)$ and $B_{\phi }={\mu }_{0}J/(2\pi r)$. The same reasoning in the primed frame gives the same expressions with primed densities.

Since $\hat{z}\times \hat{\phi }=-\hat{r}$ and $\hat{z}\times \hat{r}=\hat{\phi }$, the Lorentz transformation of electromagnetic fields gives $E_r^{\prime }=\gamma (E_r-v{B}_{\phi })$ and $B_{\phi }^{\prime }=\gamma (B_{\phi }-v{E}_r)$. Substituting the laboratory fields reproduces the primed-density formulas exactly; parallel components vanish in both frames.

The incoming and outgoing coefficients give the scattering matrixIt obeys $S(-k)=S(k{)}^{-1}$ and $S(k^{\ast }{)}^{\ast }=S(k{)}^{-1}$, so $|S(k)|=1$ for real $k$, as required by elastic unitarity. With $S=e^{2i\delta }$, the scattering phase shift satisfies $\tan \delta =k(\kappa +\alpha )/(k^2-\kappa \alpha )$. The inverse tangent needs a continuous branch: taking $\delta (\infty )=0$ gives $\delta (0)=\pi$. Equivalently $\delta =\arctan (\kappa /k)+\arctan (\alpha /k)$ for $k>0$.

The scattering length follows from $k\cot \delta =-1/a_s+O(k^2)$:For a putative bound state put $k=i\eta$, $\eta >0$. The incoming term grows as $e^{\eta r}$, so its coefficient must vanish. The incoming Jost function is $F(k)=g(-k)=(k-i\kappa )/(k+i\alpha )$; its only upper-half-plane zero is $i\kappa$. Under the regular-potential assumption this gives one bound state, with $\boxed{E=-{\hslash }^2{\kappa }^2/(2m).}$ The pole at $i\alpha$ in $S$ instead comes from the outgoing numerator: it is a redundant pole of a scattering matrix, not a second normalizable state. The S-matrix alone, without its incoming/outgoing analytic structure, would tempt an incorrect count of two states.

For completeness, the nonsingular Eckart realization of this Jost function requires $\alpha >\kappa$, as discussed in dipot.ulb.ac.be/dspace/bitstream/2013/373712/4/2306.12216.pdf. The supplied positive constants alone do not establish such a realization in every ordering; a coincident pole is likewise not evidence for two states.

There is no positive-energy scattering resonance. Indeed ${\delta }^{\prime }(k)=-\kappa /(k^2+{\kappa }^2)-\alpha /(k^2+{\alpha }^2)<0$, and the s-wave total partial-wave total scattering cross-section is $4\pi (\kappa +\alpha {)}^2/[(k^2+{\kappa }^2)(k^2+{\alpha }^2)]$, strictly decreasing for $k>0$. All poles are on the imaginary axis, with no oscillatory decaying resonance pole.

Insert the position and momentum resolutions of the identity to obtain the unitary Fourier transformThe kinetic term is multiplication by $p^2/(2m)$, and the potential has momentum kernel $⟨p|V|p^{\prime }⟩=(2\pi \hslash {)}^{-1}\int V(x)e^{-i(p-p^{\prime })x/\hslash }dx$. Thus the Schrödinger equation becomesFor the attractive delta potential, $V(x)=-({\hslash }^2\lambda /m)\delta (x)$, the kernel is the constant $-\hslash \lambda /(2\pi m)$, giving the stated positive right-hand side. For $\tilde{\psi }=N/(p^2+{\alpha }^2)$ with $\alpha >0$, constancy of the left side requires $E=-{\alpha }^2/(2m)$. Since $\int dp/(p^2+{\alpha }^2)=\pi /\alpha$, its coefficient equation givesFinally, transforming $\sqrt{\lambda }{e}^{-\lambda |x|}$ gives $\tilde{\psi }=2{\lambda }^{3/2}{\hslash }^2/[\sqrt{2\pi \hslash }(p^2+{\hslash }^2{\lambda }^2)]$, verifying the same momentum-space form and energy.