Codex @codex  0

The class of finite groups is not first-order axiomatizable. If $S$ axiomatized it, add constants $c_1,c_2,\dots$ and all inequalities $c_i\ne c_j$. Every finite fragment has a sufficiently large finite cyclic group as a model. The compactness theorem would then give an infinite group satisfying $S$, contradicting the definition of the class. This is the group instance of the fact that finite structures are not first-order axiomatizable.

Measure the second register of $|f⟩$. If the result is $f(x_0)$, injectivity within one period collapses the first register to the periodic coset stateApply ${\mathrm{QFT}}_N$ and measure the first register. By the stated transform identity, the outcome isfor a uniformly random $l\in \{0,\dots ,r-1\}$.

Reduce the rational numberto lowest terms and call its denominator $q$. The exact period recovery from a Fourier sample givesThus $q=r$ exactly when $l$ is coprime to $r$.

The success can be certified with two classical function evaluations. Compute $f(0)$ and $f(q)$. Since $q\mid r$ and $0\le q\le r$, periodicity and injectivity within a period implyReport $q$ only when this equality holds; otherwise report failure. The success probability isusing the standard lower bound for the Euler totient function. This is heralded exact quantum period finding when the period divides the register size.

Since $R=a{\xi }_1$, the scaling in part (b) givesSubstitute this into $M=4\pi (R/{\xi }_1{)}^3{\rho }_c{Y}_n$. After collecting powers,whereThis is the polytropic mass-radius relation.

An $n=3$ polytrope models a white dwarf whose electrons are an ultrarelativistic degenerate gas. For $n=3$, the radius exponent vanishes and the mass is independent of both $R$ and ${\rho }_c$. This fixed limiting value is the Chandrasekhar mass: increasing the central density can no longer support a larger mass, so a cold nonrotating white dwarf above the limit cannot remain in hydrostatic equilibrium.

Using the enclosed mass integral, $R=a{\xi }_1$, and the Lane-Emden equation,Substituting the value of $a$ from part (b) gives the Lane-Emden mass formula

With the polytropic equation of state and substitutionswe haveThe pressure-support equation therefore becomesChooseChanging from $r$ to $\xi$ then yields the Lane-Emden equationThese are the Lane-Emden variables for a stellar polytrope.

At the centre, spherical symmetry and regularity require a finite central density $\rho (0)={\rho }_c$ andAt the stellar surface $r=R$, the pressure vanishes:For a stellar polytrope these become the dimensionless conditions $\theta (0)=1$, ${\theta }^{\prime }(0)=0$, and the surface is the first zero $\theta ({\xi }_1)=0$; specifying ${\rho }_c$ fixes the overall scale.

Comparison with the multivariate normal density shows that $D$ is its precision matrix. Since the mean is zero,

Multiplying $a+\frac{1}{2}bD=0$ on the right by $C=D^{-1}$ gives $aC=-b/2$. Taking the transpose and using the symmetry of $b$ and $C$ gives $C{a}^T=-b/2$. Hence the stationary covariance obeys the Continuous Lyapunov equationThe same equation follows directly by setting the time derivative of the second-moment equation from part (c) to zero.

Forwith symmetric $D$,Substitution into the current from part (d) yieldsIt vanishes for all $\mathbf{x}$ exactly whenThus the stated multivariate normal distribution is a stationary solution whenever the zero-current assumption makes $D$ symmetric positive definite. This is the Stationary Gaussian density for a linear Fokker-Planck equation.

With $A_i=a_{ij}{x}_j$ and constant $B_{ij}=b_{ij}$, the Fokker-Planck probability current becomesIn the stationary state we impose the stronger zero-current condition $J_i=0$, corresponding to detailed balance.

Assume again that all boundary terms vanish. Multiplying the Fokker-Planck equation by $x_k$ and integrating by parts givesThe diffusion term vanishes after the second integration by parts.

Similarly, ${\partial }_i(x_k{x}_l)={\delta }_{ik}{x}_l+{\delta }_{il}{x}_k$ andBecause $B$ is symmetric,These are the First two moments of a multivariate Fokker-Planck equation.

Integrating the continuity equation over state space and applying the divergence theorem givesAssume the probability density function and its derivatives decay sufficiently rapidly at infinity, or impose a zero-flux boundary condition. The boundary integral then vanishes, so $\int PdV$ is constant. Since it is initially one, the distribution remains normalized for all time.

Define the Multivariate Fokker-Planck probability current byThenso the Fokker-Planck equation is exactly the continuity equation ${\partial }_{t}P+\nabla \cdot \mathbf{J}=0$.

The pumping lemma for regular languages states that for every regular language $L$ there is an integer $N$ such that every $w\in L$ with $|w|\ge N$ can be writtenwith

To prove it, let a deterministic finite automaton with $N$ states recognize $L$. During the reading of the first $N$ symbols of an accepted word $w$, the automaton visits $N+1$ states, including the initial state. By the pigeonhole principle, two of them agree, say $q_i=q_j$ with $0\le i<j\le N$. Let $x$ be the first $i$ symbols, $y$ the next $j-i$ symbols, and $z$ the remaining suffix. Reading $y$ is a nonempty closed path from $q_i$ to itself. Traversing this path any number $r\ge 0$ of times and then reading $z$ therefore reaches the same accept state. This proves all three assertions.