If update maps preserve a finite partial order with a bottom and top state, it suffices to run those two extremal states from the same backward horizon. Equality of their time-zero images forces equality of all initial states by the order sandwich. For ferromagnetic Ising heat-bath updates, use the same updated vertex and uniform variable in every chain; the conditional probability of $+1$ increases with every neighbour's spin. A block visiting all vertices once with all uniforms below the least conditional probability of $+1$ maps every state to all plus, proving almost-sure termination.

For a spherically symmetric bound state, write $\psi =e^{-br}\sum c_j{r}^j$. The radial Coulomb equation gives the displayed recurrence. Choosing $b=a/(2N)$ terminates it after degree $N-1$, producing a nonzero continuous polynomial times an exponentially decaying function. Its three-dimensional norm $4\pi {\int }_0^{\infty }{r}^2|\psi {|}^{2}dr$ is finite and can be normalized. The energy is $-{\hslash }^2{a}^2/(8m{N}^2)$. Continuity at the origin is compatible with the Coulomb cusp; no differentiability as a Cartesian radial function at the origin is asserted.

This is cotype of a Banach space with $q=2$. The Lp space $L^1$ has cotype 2 with constant at most $\sqrt{2}$: the triangle inequality for the Euclidean norm gives $(\sum \Vert f_j{\Vert }_1^2{)}^{1/2}\le \int (\sum |f_j(t){|}^2{)}^{1/2}dt$, and the sharp Rademacher second-moment inequality, followed by Fubini theorem, bounds this by $\sqrt{2}\mathbb{E}\Vert \sum {\epsilon }_j{f}_j{\Vert }_1$.

In a symmetric equal mixture of two ordered pure thermodynamic phases with magnetizations $\pm M_0$, the mean spin vanishes but the large-distance two-spin expectation tends to $M_0^2$. Thus the connected correlation function of the mixture has a nondecaying part. The correlation-function susceptibility sum rule gives an extensive contribution ${\beta }_{\mathrm{th}}N{M}_0^2$ to susceptibility per spin. This is phase-switching response, not the connected susceptibility within one pure phase. The latter uses the thermodynamic limit followed by a one-sided zero-field limit, and subtracts the nonzero pure-phase mean before defining the correlation length.

If $\sqrt{2}<s\le 2$, every nondegenerate subinterval of $A=[1-s,1]$ eventually maps onto $A$. Away from the critical point, lengths multiply by $s$; one fold loses at most a factor of two. If an interval and its image both contain zero, its second image is $A$. Otherwise every two steps increase length by at least $s^2/2>1$, which cannot continue inside bounded $A$. Applying the intermediate value theorem to an interval that covers itself then gives a periodic point in every subinterval.

// Target: dynamical-systems.bigb

For $r\ge 2$, the core is zero when $n<r$, the nonnegative unit simplex when $n=r$, the single allocation $x_i=1/r$ when $n\ge 2r$ is divisible by $r$, and empty for all other $n\ge r$. To prove the obstruction, every $r$-person coalition requires payoff at least one. Summing these constraints counts each player $\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ times and gives $r\lfloor n/r\rfloor /n\ge 1$, impossible when $r\nmid n$. For a divisible $n$, all such coalition sums must equal one. Comparing coalitions with the same $r-1$ members makes all individual payoffs equal when $n>r$. The uniform allocation satisfies every coalition constraint because $|S|/r\ge \lfloor |S|/r\rfloor$. At $n=r$ all proper coalition values are zero, giving the whole simplex instead.

An FFT butterfly combines corresponding outputs of two half-sized transforms using a twiddle factor $w$. Sharing the product $wb$ costs one complex multiplication and two additions. For the unnormalized positive-exponent inverse discrete Fourier transform of length $2m$, use $w=e^{2\pi i\ell /(2m)}$ at index $\ell$.

For a taxi starting at any vertex of a rectangle with side lengths $s,t$, a uniform pickup vertex gives approach-distance law uniform on $0,s,t,s+t$. Conditional on that pickup, a uniform distinct destination gives passenger-distance law uniform on $s,t,s+t$, independent of the pickup vertex. Hence the two distance variables are independent despite sharing that vertex. Their convolution has twelve equally likely pairs. Its mean is $7(s+t)/6$ and its variance is $(s^2+t^2)/4+2(s^2+t^2-st)/9$. Repeated numerical sums combine their multiplicities when some distances coincide.

Choose a uniform random ordering of $N$ and restrict it to $T$. The restriction is uniform, and each player's predecessor set in $T$ is contained in their predecessor set in $N$. Increasing marginal contributions in a convex cooperative game make the full-population contribution at least the restricted contribution. Averaging proves the displayed Shapley value inequality. Summing over $i\in T$ and using efficiency in the restricted game proves every core inequality.

For $n$ points in a triangle and two fixed endpoint vertices, the longest convex chain length is the largest number of sampled points in a convex chain joining those endpoints. It is a certifiable function: a chain of length $a$ supplies a certificate of exactly $a$ point coordinates. Changing one point changes the maximum by at most one, since deleting that point from any optimal chain loses at most one vertex. For independent uniform points, its median is $O(n^{1/3})$ and its fluctuation scale about the median is at most $n^{1/6}$, as proved by convex-chain probability in a triangle and convex-chain median concentration.

For independent uniform points in $0\le y\le x\le 1$ with the two diagonal endpoints, the displayed formula is exact. Order the $k$ intermediate points so that both coordinates increase, and write the $k+1$ positive coordinate spacings as $(u_i)$ and $(v_i)$, with each sum equal to one. The two spacing vectors have uniform simplex measure, independently, and the full ordered-coordinate region has volume $1/(k!{)}^2$. Simultaneous permutation of the spacing pairs preserves that measure, so all $(k+1)!$ orders of the ratios $v_i/u_i$ have equal volume. Strictly increasing ratios characterize a convex chain and automatically place the intermediate points below the diagonal. Thus its ordered volume is $1/((k!{)}^2(k+1)!)$. There are $k!$ ways to assign labels, and uniform triangle density is $2^k$, giving the formula. Consequently the first moment method bounds the probability of any length-$k$ chain among $n$ points by $\left(\genfrac{}{}{0ex}{}{n}{k}\right)2^k/(k!(k+1)!)$.

For nonnegative integrable functions converging almost everywhere to an integrable $f$, convergence of their integrals implies L1 convergence. Indeed $\min (f_n,f)\to f$ almost everywhere and is dominated by $f$, so the dominated convergence theorem gives convergence of its integral. Use $|f_n-f|=f_n+f-2\min (f_n,f)$. This applies to probability densities, giving convergence in total variation distance.

For fixed $k$ independent random variables whose families have the same large deviation principle with good rate function $I$, the product large-deviation principle gives the sum of coordinate costs. Under the minimum map, one coordinate must equal $m$ and the others must be at least $m$, yielding the displayed rate function. If $I$ is convex with unique zero $\mu$, then $J(m)=I(m)$ for $m\le \mu$ and $J(m)=kI(m)$ for $m\ge \mu$.

For $a<b$, the contraction principle for large deviations under scalar clipping gives $K(z)=I(z)$ in $(a,b)$, the displayed endpoint costs, and infinite cost outside $[a,b]$. The endpoint costs can differ from $I(a)$ and $I(b)$ because clipping collects a whole input tail into a single output atom.

For a homologous stellar polytrope of index $3/2$, take $P=K{\rho }^{5/3}+\mathcal{R}\rho T/\mu$ and zero surface pressure. The stellar virial theorem, with $\Omega =-6G{M}^2/(7R)$, givesThe normalized density profile fixes $A$ independently of mass and radius at fixed composition. Writing ${\overline{T}}_M=\alpha /R-\delta /R^2$ shows that its maximum is ${\alpha }^2/(4\delta )=3\mu G^2{M}^{4/3}/(49\mathcal{R}A)$, at $R=7A{M}^{-1/3}/(3G)$. Below this radius contraction increases the fraction of support provided by electron degeneracy pressure and lowers the thermal temperature. A fixed normalized profile also fixes the ratio of central to mass-averaged temperature, giving a hydrogen-burning minimum mass.

For a sector of width $\varphi$ and an independent uniformly distributed central bearing, a specified bearing is covered with probability $\varphi /(2\pi )$. Two antipodal bearings cannot both be covered when $\varphi <\pi$. For $\varphi \ge \pi$, the allowable centre arcs overlap in two components of length $\varphi -\pi$, proving the displayed conditional probability. Averaging over the width's probability density function gives unconditional coverage probabilities.

A Q-matrix has nonnegative off-diagonal entries and row sums zero. Its diagonal entry is the negative total rate out of the state.

An Equilibrium distribution of a continuous-time Markov chain is a probability row vector $\pi$ invariant under its transition semigroup:It satisfies $\pi Q=0$ when the generator sums are defined. An invariant law is a property of the entire evolution, so formal generator equations must be interpreted with the process's nonexplosion and domain conditions.

For a finite irreducible chain, the equilibrium distribution is unique and every transition row converges to it. For a general finite chain, the limiting row from state $i$ is a mixture of the stationary distributions ${\pi }^C$ of its closed irreducible classes:The chain eventually enters one of these classes and then mixes within it. If every closed class is a single absorbing state, this reduces to absorption probabilities at those states. Merely having some absorbing states does not exclude other closed classes.

Holding times explain the link to an embedded jump chain. For a finite irreducible chain with strictly positive exit rates $q_i$ and jump-chain Stationary distribution $\alpha$, the time-stationary distribution isVisit frequency is multiplied by mean time $1/q_i$ per visit. In the countable setting, these time weights need a finite positive normalizer and a nonexplosive process before they define an equilibrium probability. For an alternating two-state jump chain, $\alpha =(1/2,1/2)$ and rates $a,b$ give $\pi =(b/(a+b),a/(a+b))$: slower holding rates increase time occupancy.

A continuous-time Markov chain with finite holding rates is nonexplosive when its jump times have no finite accumulation almost surely. It then remains in its specified state space at every finite time, rather than being lost to a cemetery state after infinitely many jumps. Bounded rates are sufficient for nonexplosion, but finite rates at individual states alone are not.

Consider states $C_i,W_i$ for integer levels $i\ge 0$ and strictly positive rates $\lambda ,\mu ,\alpha ,\beta$. From $C_i$ the chain moves to $C_{i+1}$ at rate $\lambda$ and to $W_i$ at rate $\alpha$. From $W_i$, $i\ge 1$, it moves to $W_{i-1}$ at rate $\mu$ or to $C_i$ at rate $\beta$. At the reflecting boundary $W_0$ there is only the switch to $C_0$; no negative level is allowed. All rates are bounded, so the chain is nonexplosive.

Let $h$ be the probability of hitting $W_0$, with $h_{W_0}=1$. First-step equations arewith the first equation also valid at $i=0$. Put $\vartheta =\mu (\lambda +\alpha )/[\lambda (\mu +\beta )]$. If $\mu \alpha <\lambda \beta$, then $\vartheta <1$ andSubstitution verifies both equations and the boundary. Eliminating the $C$ variables gives a second-order recurrence with modes $1,{\vartheta }^i$. The displayed decaying solution is the minimal nonnegative harmonic solution and hence the Hitting probability.

When $\mu \alpha \ge \lambda \beta$, the growing mode is excluded by boundedness; at equality the second mode is linear and is likewise excluded. The only bounded solution with the boundary value is one, proving almost-sure hitting. Since the specified reflected chain is irreducible, these cases classify it as recurrent or transient respectively. Specifying the boundary rule is essential for that recurrence statement; making $W_0$ absorbing would be a different chain.