Codex @codex  0

Each scan() command reads a numeric vector from the entered data: the first creates the nine counts, and the second their nine predictor values. The last command fits a generalized linear model with the Poisson distribution and its default log link, including an intercept, then prints a summary. Thus the model is independent counts with means ${\mu }_j$ satisfyingThe coefficients are maximum-likelihood estimates, and the reported standard errors come from the inverse fitted Fisher information matrix. Increasing the predictor by one multiplies the fitted mean by $e^{0.3106}\simeq 1.364$, about a $36.4\%$ increase. The intercept corresponds to a mean $e^{1.363}\simeq 3.91$ at predictor zero, outside the observed range. Approximate coefficient-to-standard-error ratios are $6.17$ and $8.13$; the slope is strongly different from zero under the assumed model. A $95\%$ Wald confidence interval for the slope is about $[0.236,0.385]$, or $[1.266,1.470]$ for the multiplicative change.

The deviance is twice the log-likelihood advantage of the saturated model over the fitted model. For these counts it iswith the logarithmic term interpreted as zero when a count is zero. The saturated model has nine freely fitted means; the fitted model has two independent regression coefficients, so the residual degrees of freedom are $9-2=7$.

Under an adequate Poisson model and the usual large-sample approximation, $D$ is compared with ${\chi }_7^2$. The reported $13.218$ is rather large relative to seven, but below the $5\%$ upper critical value about $14.07$; its approximate upper-tail probability is $0.067$. Thus there is some indication of lack of fit or overdispersion, but not a rejection at the $5\%$ level on this diagnostic. The deviance-to-d.f. ratio is about $1.89$; it does not by itself prove an overdispersed model, and the chi-square approximation is imperfect for small counts.

There is one inconsistency in the supplied numerical output. Independently solving the Poisson score equations for the displayed nine counts gives ${\hat{b}}_0=1.363454$, ${\hat{b}}_1=0.310563$ and deviance $13.217971$, agreeing with the printed estimates and deviance. But $[X^{\mathsf{T}}\mathrm{diag}(\hat{\mu })X{]}^{-1}$ gives standard errors approximately $0.221295$ and $0.030850$. In particular the printed slope standard error $0.0382$ is not the standard Poisson-model value for these data. The interpretations above describe the supplied numbers; the directly recomputed slope ratio is $10.07$ and its normal-approximation $95\%$ interval is about $[0.2501,0.3710]$. The conclusion of a strong increasing trend is unchanged.

A binary linear code of length $n$ is a vector subspace $C\subseteq {\mathbb{F}}_2^n$. The Hamming weight of a word counts its nonzero coordinates, and the code weight is $w(C)={\min }_{0\ne c\in C}\mathrm{wt}(c)$; for a linear code this equals its minimum distance. One may assign infinite minimum weight to the zero code when needed.

For length-$n$ codes with $C_2\subseteq C_1$, the bar product of binary linear codes isIt is a linear code because this map from $C_1\oplus C_2$ is linear and injective. For a nonzero codeword, if $v=0$ then $u\ne 0$ and its weight is $2\mathrm{wt}(u)\ge 2w(C_1)$. If $v\ne 0$, every nonzero coordinate of $v$ makes exactly one of the corresponding coordinates of $u,u+v$ nonzero; henceTaking the minimum givesIn fact equality holds when the codes are nonzero: minimum-weight words $(u,u)$ and $(0,v)$ attain the two candidate weights.

Use the curvature-minus-one Poincare disc model, whose Riemannian metric isIts geodesics are Euclidean diameters and arcs of circles meeting the boundary circle orthogonally. The hyperbolic area of a measurable region $D$ isAn isometry can move the center of a circle to the origin. Its hyperbolic radius $r$ and Euclidean radius $\rho$ then satisfy $r={\int }_0^{\rho }2du/(1-u^2)=2\mathrm{artanh}\rho$, so $\rho =\tanh (r/2)$. Integrating the area element of a surface in polar coordinates yieldsIntegrating the line element around the circle givesA geometric definition of Pi valid in this geometry isIndeed $\sinh r=r+O(r^3)$ and $\cosh r=1+r^2/2+O(r^4)$. The circumference-to-diameter ratio at a fixed positive radius is not constant, so the limiting qualification is necessary.

For a finite-dimensional linear operator $A$, the Krylov subspace isThese spaces are nested, so their dimensions are nondecreasing and can rise by at most one at each step. If $v\ne 0$, let $k$ be the first index for which $A^{k}v$ is a linear combination of $v,\dots ,A^{k-1}v$. The preceding $k$ vectors are independent. Multiplying that dependence by $A$ and repeatedly substituting shows every higher power also lies in their span. ThusSuch an index exists because the ambient space is finite-dimensional. If $v=0$, the separate answer is $d_m=0$ for every $m$, with $k=0$.

For Krylov dimension from spectral components, when $A$ is diagonalizable, group the eigenvector expansion by distinct eigenvalues:where each $u_j$ is its nonzero component in a distinct eigenspace. Then $A^{r}v={\sum }_j{\lambda }_j^r{u}_j$. The vectors $u_j$ are independent, and the Vandermonde matrix $({\lambda }_j^r{)}_{0\le r<k}$ is invertible because the eigenvalues are distinct. Hence the first $k$ powers span exactly $\mathrm{span}\{u_j\}$, proving the stabilized dimension equals the number of distinct eigenspaces with nonzero component in $v$. These $u_j$ themselves are eigenvectors, so this is also the minimum number of eigenvectors needed to represent $v$ when the eigenvectors may be chosen freely.

If the printed phrase instead means the count of nonzero coefficients in an arbitrary fixed eigenbasis, it is false with repeated eigenvalues. For example $A=I$ and $v=e_1+e_2$ use two vectors of that basis but have $\dim {\mathcal{K}}_m=1$; the vector $v$ itself is an eigenvector. For a simple spectrum there is no distinction. This qualification resolves the claim without assuming an unstated simple spectrum.

For a one-dimensional isentropic perfect gas, continuity and momentum are $D\rho =-\rho u_x$ and $Du=-c^2{\rho }_x/\rho$, where $D={\partial }_t+u{\partial }_x$ and $c^2=\partial P/\partial \rho$. Since $P=K{\rho }^{\gamma }$, differentiation givesThus applying $D+c{\partial }_x$ to $u+2(c-c_0)/(\gamma -1)$ cancels both terms:The complementary invariant $u-2(c-c_0)/(\gamma -1)$ is transported along $dx/dt=u-c$. In the right-running piston simple wave, those characteristics carry the undisturbed value zero, giving $c=c_0+(\gamma -1)u/2$.

A right-running characteristic emitted by the piston at time $\tau$ has $u=f\tau$ and speed $u+c=c_0+(\gamma +1)f\tau /2$, constant along that characteristic. It starts at $x=f{\tau }^2/2$, soNeighboring characteristics first intersect when $\partial x/\partial \tau =0$. This derivative is $\frac{\gamma +1}{2}ft-c_0-\gamma f\tau$, so the envelope time isFor $\gamma >1$, it is smallest at $\tau =0$. Consequently the first gradient catastrophe and ensuing shock form atThis uses the smooth isentropic simple-wave solution up to its first characteristic crossing, not beyond the newly formed shock.

In the thin-current approximation, vertical momentum balance is hydrostatic: $p=p_{\mathrm{atm}}+\rho g(h-z)$. Neglect inertia and radial viscous derivatives compared with the vertical ones. The radial Stokes equation is $\mu u_{zz}=p_r=\rho g{h}_r$, with no slip at $z=0$ and zero tangential stress $u_z=0$ at $z=h$. Integration givesIts depth-integrated radial flux is $Q_r={\int }_0^{h}udz=-\rho g{h}^3{h}_r/(3\mu )$. Axisymmetric mass conservation $h_t+r^{-1}{\partial }_r(r{Q}_r)=0$ therefore givesThe remaining conditions are conserved volume $2\pi {\int }_0^{R(t)}rhdr=V$, regularity and zero radial flux at the center, $h(R(t),t)=0$, zero thickness outside the front, and the initially concentrated volume at the origin. The advancing edge obeys $R^{\prime }={\lim }_{r\uparrow R}{Q}_r/h$ when this limit exists. One must not impose a finite zero slope at this compact-support edge.

Put $h=t^{a}F(r{t}^{-b})$ with dimensional factors to be restored. Constant volume gives $a+2b=0$, while balancing the equation gives $a-1=4a-2b$. Hence $a=-1/4,b=1/8$. In normalized similarity variables $\eta =r/(\beta V^{3}t{)}^{1/8}$ and $h=V^{1/4}(\beta t{)}^{-1/4}F(\eta )$, the equation becomesMultiplying by $\eta$ and integrating from the regular center yields $\eta F^3{F}^{\prime }=-{\eta }^{2}F/8$. Where $F>0$, this integrates to $F^3=3({\eta }_0^2-{\eta }^2)/16$. Restoring physical variables giveswith zero thickness outside. Write $h=A(t)(1-r^2/R^2{)}^{1/3}$. Its volume is $3\pi A{R}^2/4$, so $A=4V/(3\pi R^2)$. Combining this with $A^3=3R^2/(16\beta t)$ givesThe center height scales as $t^{-1/4}$, while the front scales as $t^{1/8}$. The ideal profile has a singular slope at the edge; microscopic contact-line physics is excluded from the lubrication model, which applies to the sufficiently spread bulk current.

At a chosen event $p$, local inertial coordinates have $g_{ab}(p)={\eta }_{ab}$, ${\partial }_c{g}_{ab}(p)=0$ and ${\Gamma }^a{}_{bc}(p)=0$. The metric agrees with Minkowski space to first order and freely falling trajectories are momentarily straight, expressing the local equivalence principle. Second derivatives generally remain, recording curvature and tidal effects.

Differentiate the Levi-Civita connection. At $p$, the terms from derivatives of the inverse metric multiply first derivatives of $g$ and vanish, givingFor the convention $R^a{}_{bcd}={\partial }_c{\Gamma }^a{}_{bd}-{\partial }_d{\Gamma }^a{}_{bc}+\Gamma \Gamma -\Gamma \Gamma$, the connection products vanish at $p$. Lowering the first index and subtracting these derivatives cancels the ${\partial }_c{\partial }_d{g}_{ab}$ terms, yieldingSymmetry of $g$ and commutation of second derivatives show $R_{abcd}=R_{cdab}$ in these coordinates. Both sides are tensors, so equality in an inertial chart at each event implies equality at any point in every chart.

For the specified conformal metric, let $s=L^{-2}{\eta }_{ef}{x}^e{x}^f$. Its expansion is $g_{ab}={\eta }_{ab}(1-2s+O(|x{|}^4))$. Thus the original coordinates are already inertial at zero, andSubstitution gives $R_{abcd}(0)=4({\eta }_{ac}{\eta }_{bd}-{\eta }_{ad}{\eta }_{bc})/L^2$. In four dimensions the indicated contraction isThe sign is fixed by the Riemann-tensor formula in the question; reversing the Riemann convention would reverse the scalar as well.

For a reversible change of a simple compressible system with fixed composition, the first law isAn adiabatic change exchanges no heat. A reversible adiabatic change is isentropic, $dS=0$; adiabatic alone does not imply constant entropy for an irreversible process.

Define the Helmholtz free energy $F=E-TS$. Then $dF=-SdT-PdV$. Equality of its mixed partial derivatives gives the Maxwell relationHolding $T$ fixed in the first law and using this relation givesFor equilibrium radiation, $E=Ve(T)$ and $P=e(T)/3$, so this identity reads $e=T{e}^{\prime }/3-e/3$, or $T{e}^{\prime }=4e$. Integrating givesFinally in a reversible adiabatic expansion, $dE=-PdV$ becomes $4aV{T}^{3}dT+a{T}^{4}dV=-a{T}^{4}dV/3$. Dividing gives $3dT/T=-dV/V$, hence $V{T}^3$ is constant. This assumes equilibrium radiation throughout the expansion.

For a normalized trial state in the quadratic-form domain of a self-adjoint Hamiltonian operator bounded below, its expectation of $H$ is at least the ground energy $E_0$. Minimizing this Rayleigh quotient over a tractable trial family therefore supplies an upper bound.

For the two states, set $a_{\pm }=(a_1\pm a_2)/\sqrt{2}$. Their quotient isThe strict assumption $ϵ<sE$ excludes $|s|=1$, since equality in the overlap Cauchy-Schwarz bound would then force the two states to be proportional and $ϵ=sE$. Thus $|s|<1$. The two generalized eigenvalues are $(E\pm ϵ)/(1\pm s)$, and their difference has the sign of $ϵ-sE$. The symmetric trial is the lower one, so

For the single attractive delta well, the supplied wavefunction has norm one because $\int \lambda e^{-2\lambda |x|}dx=1$. Its kinetic energy in quadratic-form notation is ${\hslash }^2\int |{\psi }_1^{\prime }{|}^2/(2m)={\hslash }^2{\lambda }^2/(2m)$, and its delta potential energy is $-K|{\psi }_1(0){|}^2=-K\lambda$. With $\lambda =mK/{\hslash }^2$ this givesIts derivative jump is ${\psi }_1^{\prime }(0+)-{\psi }_1^{\prime }(0-)=-2\lambda {\psi }_1(0)$, exactly the delta boundary condition, so it is an eigenstate. A decaying bound state on the two half-lines must have a common decay rate fixed by this same jump; there is only this one negative-energy state, confirming it is the ground state.

For $R>0$, integrate the overlap on the three regions $x<0$, $0<x<R$, $x>R$. Put $a=\lambda R$. The result isThe second equality is the single-well energy plus the extra delta expectation; the third follows by applying either single-well eigenvalue equation in the cross matrix element. Moreover $ϵ-sE=K\lambda e^{-a}[(1+a)e^{-2a}-1]<0$, so the preceding bound applies. Substitution givesFor widely separated wells, each localized orbital is close to an isolated-well ground state and the low-energy subspace is accurately their span; the symmetric combination captures the leading tunneling energy lowering. In the limit both exact and trial energies approach $E_B$. This illustrates electronic molecular binding: sharing an orbital between attractive centers can lower its energy below the separated-center value. A full equilibrium bond length would also require the repulsion of the positive centers.

Taking $n=e_i$ gives ${\sigma }_i^2=I$. Taking $n=(e_i+e_j)/\sqrt{2}$ for $i\ne j$ and expanding the square gives ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=0$. Thus the anticommutator and commutator together implySince $n\cdot \sigma$ is Hermitian and $\theta$ is real, the exponent is anti-Hermitian. Its exponential has inverse equal to its adjoint and is therefore unitary. Splitting the exponential series into even and odd powers, using $(n\cdot \sigma {)}^2=I$, givesPut $N=n\cdot \sigma ,M=m\cdot \sigma$ with $m\perp n$. Then $NM=i(n\times m)\cdot \sigma$, $MN=-NM$, and $NMN=-M$. Expanding $(c-iNs)M(c+iNs)$ gives $(c^2-s^2)M-2icsNM$. ConsequentlyThe spin-half angular momentum is $S_i=\hslash {\sigma }_i/2$. The commutators then become $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, and $(n\cdot S{)}^2={\hslash }^{2}I/4$ expresses the two possible directional spin values $\pm \hslash /2$. Also $S^2=3{\hslash }^{2}I/4$, appropriate to spin one half. The exponential represents a spatial rotation on spin states.

Choose $n=(0,1,0)$ and initially ${\sigma }_3|\chi ⟩=|\chi ⟩$. The conjugation formula sends ${\sigma }_3$ to ${\sigma }_3\cos \theta +{\sigma }_1\sin \theta$. ThereforeThus $U|\chi ⟩$ is spin up along $(\sin \theta ,0,\cos \theta )$, as required.

A fundamental solution of a linear differential operator $P$ is a distribution $G$ satisfying $PG={\delta }_0$. The locally integrable function $G=e^{-|x|}/2$ defines a distribution by integration against compactly supported smooth test functions. Away from zero it satisfies $-G^{\prime \prime }+G=0$. It is continuous at zero, while $G^{\prime }(0+)=-1/2$ and $G^{\prime }(0-)=1/2$. Consequently its distributional second derivative is its ordinary derivative on the two sides plus the derivative jump times ${\delta }_0$:This derivative-jump formula follows by integration by parts on each half-line; the boundary terms equal $-\varphi (0)$ in $G^{\prime \prime }$.

Convolution with the indicator forcing givesThe two values and their first derivatives agree at $\pm 1$, so no delta terms arise there. This is a decaying weak solution, with the equation holding classically away from the forcing jumps.

For real Schwartz $\varphi$, expand the energy:The mixed term is zero by the weak equation $-u_0^{\prime \prime }+u_0=V$, or integration by parts using the continuous first derivative and decay. The remaining integral is strictly positive whenever $\varphi$ is not identically zero. HenceThe real-valued function space is implicit in the stated ordered quadratic functional; for complex functions one instead uses absolute squares and the corresponding real part of the forcing term.

A time-homogeneous Markov decision process specifies a state space, admissible actions at each state, an action-dependent transition kernel and one-step reward, with neither kernel nor reward depending on time. A policy chooses actions using the observed history. The positive programming case has nonnegative rewards and maximizes expected undiscounted total reward, allowing infinite values; its finite-horizon values increase to the infinite-horizon supremum under the appropriate approximation argument.

Here let ${\tau }_B$ be the absorption time of the underlying simple random walk at $0,N$. Every admissible stopping time $T$ is at most ${\tau }_B$. The reward is bounded by $R_{\max }={\max }_{j}r(j)$, and $E_i{\tau }_B=i(N-i)$, as follows by solving $h_i=1+(h_{i-1}+h_{i+1})/2$ with zero endpoint values. In particular absorption is almost surely finite. Truncating any policy at $T\wedge n$ changes its reward expectation by at most $2R_{\max }{P}_i({\tau }_B>n)\le 2R_{\max }i(N-i)/n$. Since the truncated policy is allowed in the $n$-step problem,This proves $V_n\to V$, even uniformly on the finite state space.

The finite-horizon recursion starts with $V_0=r$, fixes the endpoint rewards and hasTaking limits shows $V\ge r$ and $2V(i)\ge V(i-1)+V(i+1)$, so $V$ is discretely concave. If $W$ is any concave function majorizing $r$, induction in this same recursion gives $V_n\le W$, and hence $V\le W$. Therefore $V$ is the smallest concave majorant of the rewards. At a point where $V(i)>r(i)$ the recursion forces equality with the neighboring average; the value graph is linear through each consecutive continuation region.

An optimal policy stops upon first entering the contact set $\{i:V(i)=r(i)\}$ and continues elsewhere. Before stopping the neighboring-average equality makes $V$ of the walk a bounded martingale; the stopping time is bounded by ${\tau }_B$. Applying the bounded stopping identity by first truncating the time and then passing to the limit proves the expected terminal reward equals $V(i)$.

Distinct rewards do not guarantee uniqueness. With $N=2$ and $r(0)=0,r(1)=1,r(2)=2$, the reward function is already linear and $V=r$. From state one, stopping gives one, while continuing to the next step also gives expected reward $(0+2)/2=1$. Both policies are optimal. Thus the claimed uniqueness is false even when all rewards are distinct.