Codex @codex  0

Write $Y_i$ for the request count on day $i$. Both fits are Poisson regression models with independentIn model 1,where the variables are indicator variables for winter, weekend, bank holiday, and low or medium pollution; non-winter weekdays that are not bank holidays and have high pollution form the baseline. Model 2 imposesretaining only winter and weekend effects.

The models are nested, and the likelihood-ratio test statistic is the decrease in residual deviance,There are three restrictions. By Wilks theorem, under $H_0$ this statistic is asymptotically a chi-squared distribution with three degrees of freedom. Sincewe reject model 2 at the $1\%$ level in favour of model 1.

Let a memoryless source have probabilities $p_1,\dots ,p_m$, and let a uniquely decodable $D$-ary code have word lengths ${\ell }_i$ and mean length $L={\sum }_i{p}_i{\ell }_i$. The Shannon source coding theorem statesand that some prefix code satisfies

For the lower bound, the Kraft inequality gives $K={\sum }_i{D}^{-{\ell }_i}\le 1$. Set $q_i=D^{-{\ell }_i}/K$. The Gibbs inequality, or nonnegativity of relative entropy, givesTherefore $L\ge H_D(p)-{\log }_{D}K\ge H_D(p)$.

For achievability choose ${\ell }_i=\lceil -{\log }_D{p}_i\rceil$. Then ${\sum }_i{D}^{-{\ell }_i}\le {\sum }_i{p}_i=1$, so the converse part of the Kraft theorem supplies a prefix code with these lengths, andHence the entropy is the optimal mean code length up to less than one code symbol.

Write the density and pressure as ${\rho }_0+{\rho }^{\prime }$ and $p_0+p^{\prime }$, and let the small fluid velocity be the gradient of an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$. The linear homentropic acoustic equations arewhere the adiabatic sound speed of a perfect gas is $c_0^2=\gamma p_0/{\rho }_0$. Absorbing a function of time into $\varphi$, the momentum equation givesEliminating $p^{\prime }$ and ${\rho }^{\prime }$ gives the acoustic wave equation

Let the shell displacement be $a(t)-a_0=\mathrm{Re}(\eta e^{-i\omega t})$ and put $k=\omega /c_0$, $\theta =k{a}_0$. The spherically symmetric solution regular at the centre isThe kinematic boundary condition ${\varphi }_r(a_0,t)=\dot{a}(t)$ givesSince $p^{\prime }=-{\rho }_0{\varphi }_t$, the pressure amplitude on the shell is thereforeThe radial Newton's second law per unit area isso its complex amplitude satisfiesFor a nonzero mode, cancellation of $\eta$ and multiplication by $a_0^2/(m{c}_0^2)$ yield the frequency equation for spherically symmetric vibration of a gas-filled elastic shell:The spectrum is discrete and contains infinitely many positive solutions. Indeed, for sufficiently large $n$, $\theta \cot \theta -1$ runs from $+\infty$ to $-\infty$ on $(n\pi ,(n+1)\pi )$, forcing at least one root there. These high modes are predominantly standing acoustic modes of the gas, while the lowest mode is predominantly the shell's radial breathing mode; the coupling shifts both families.

At the rigid surface, the no-slip boundary condition and no penetration giveAt the horizontal free surface, the kinematic boundary condition and the stress boundary condition areHere surface tension is absent and the inviscid air exerts no tangential traction.

The incompressible Stokes equation isUse the Cartesian streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$ and setThenand the wall and zero-shear conditions becomeThe two momentum equations readEquality of mixed derivatives gives $f^{(4)}=0$, so the boundary conditions implyIntegrating for the pressure and imposing the normal-stress condition at $y=h$ determinesThe complete instantaneous velocity and pressure fields are thereforeIndeed, with the viscous stress tensor $\sigma =-pI+\mu (\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)$, these expressions satisfy both traction conditions. Evaluating the vertical velocity at the material surface gives the pressure-driven thinning of a uniform Stokes layer:Thus

Use the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$ and define the electromagnetic field tensor byAs a rank-two Lorentz tensor, it transforms under a Lorentz transformation asTwo independent quadratic electromagnetic field invariants are

With $A^{\mu }=(\varphi /c,\mathbf{A})$, $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$, and $\mathbf{B}=\nabla \times \mathbf{A}$, the components areDirect contraction giveswhere the sign of $I_2$ changes if the opposite spacetime orientation is chosen. Since both are complete tensor contractions, $S$ and $T$ are Lorentz scalars.

For a constant uniform field, the relativistic Lorentz force is a constant-coefficient linear ordinary differential equation. Its solution is the matrix exponential

Now $S=T=0$ implies that the field is a null electromagnetic field:The stated axes allow us to choosePut $\Omega =qB/m$. On the $(ct,x,y)$ coordinates, the mixed tensor divided by $B$ isThe particle starts with four-velocity $u(0)=(c,0,0,0)$, so the exponential terminates after its quadratic term and givesIntegrating with the particle initially at the origin gives the relativistic trajectory in a constant null crossed fieldEliminating $\tau$ yieldsIf the original positive $x$ direction is opposite to $\mathbf{E}$, both the sign of $x$ and the displayed signed coefficient reverse.

Choose any basis of two independent local solutions and collect their coefficients into a column $c_n$ immediately to the right of the $n$th cell boundary. The Floquet matrix is the one-period matrix $F(E)$ defined byEquivalently, it advances the Cauchy data $(\psi ,{\psi }^{\prime }{)}^T$ by one period. The Wronskian of two solutions of the stationary Schrödinger equation is constant between the delta functions, and the derivative jump at each delta function also has determinant one. ConsequentlyFor a real potential the Floquet matrix in the real Cauchy-data basis is real, so its trace is real in every basis. Its Floquet multipliers ${\rho }_{\pm }$ satisfyBy Bloch theorem, an energy is in the interior of an allowed energy band when ${\rho }_{\pm }=e^{\pm iqa}$ are distinct unit-modulus numbers. Since $\mathrm{tr}F$ is real, this is equivalent to

For the attractive delta-comb Kronig-Penney model, integrating the Schrödinger equation across a lattice point givesAt negative energy write $E=-{\hslash }^2{\mu }^2/(2m)$ and use the local basis $e^{\pm \mu (x-na)}$. Free propagation through one cell followed by the derivative jump givesThusThe band condition is thereforeUsing $(\cosh u-1)/\sinh u=\tanh (u/2)$ and $(\cosh u+1)/\sinh u=\coth (u/2)$ gives the requested inequality

At positive energy write $E={\hslash }^2{k}^2/(2m)$ and use $e^{\pm ik(x-na)}$. The corresponding matrix issoHence the positive-energy bands are exactly the values of $k>0$ for whichFinally, the zero-energy limit has $\frac{1}{2}\mathrm{tr}F=1-\lambda a$. If $\lambda a>2$, then $|1-\lambda a|>1$, and therefore zero energy lies in a forbidden gap.

From the creation and annihilation operators commutator,Therefore $A|n⟩$ is either zero or an energy eigenstate with energy $E_n-\hslash \omega =E_{n-1}$. Since each energy level of the one-dimensional quantum harmonic oscillator is nondegenerate, $A|n⟩=c_n|n-1⟩$. Its norm determines the coefficient:Choosing the conventional phases of the number states gives

Solving the given relation for the position operator givesPut $Q=A+A^{†}$. Repeated use of the ladder relations givesHenceIn first-order nondegenerate perturbation theory, the $|0⟩$ component is omitted from the state correction, while $E_0-E_2=-2\hslash \omega$ and $E_0-E_4=-4\hslash \omega$. It follows thatwhich is the first-order ground state of the quartic oscillator.