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Each state has exactly $k(n-k)$ neighbours: choose one occupied coordinate and one unoccupied coordinate to swap. The normalized proposal distribution is thereforeThe swap relation is symmetric and the number of neighbours is constant, so $q(v\mid u)=q(u\mid v)$. The discrete uniform distribution used as the prior distribution is also constant on $S_k$, with mass ${\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^{-1}$. Both factors cancel in the Metropolis–Hastings acceptance probability.

Let $Q={\Sigma }^{-1}$ and $d=v-u$. The displayed multivariate normal distribution presupposes that the covariance matrix $\Sigma$ is a positive-definite matrix, so $Q$ is a symmetric matrix. Expanding the difference of quadratic forms givesConsequentlyThe truncation at zero is essential: a log acceptance ratio can be positive, but the logarithm of an acceptance probability cannot.

For the particular proposed swap, let $a=i_1$ be the coordinate changing from zero to one and $b=i_2$ the coordinate changing from one to zero. Then $d=e_a-e_b$, and the untruncated scalar simplifies further toOnly two rows and four entries of $Q$ are needed for this evaluation. This is Metropolis sampling on fixed-cardinality subsets.

Solvent quality. The Flory interaction parameter $\chi$ measures the free-energy cost of polymer–solvent contacts relative to like contacts. In the simplest lattice model,where $z_c$ is the coordination number and the $ϵ$ values are contact energies. For lattice-site volume $v_0$, the effective two-segment excluded volume is $v_{\mathrm{eff}}=v_0(1-2\chi )$. Its sign determines the leading tendency of a long dilute flexible chain.

For $\chi <1/2$, $v_{\mathrm{eff}}>0$: a good solvent favors a swollen coil because segment encounters are effectively repulsive. For $\chi =1/2$, the leading attraction and repulsion cancel, giving approximately ideal chain statistics. For $\chi >1/2$, effective attraction favors contraction in a poor solvent, eventually producing a compact polymer globule under sufficiently poor conditions. This does not mean that every finite-chain solution with $\chi >1/2$ is automatically phase-separated; that also depends on composition and chain length.

The condition $\chi =1/2$ is the theta temperature condition for that polymer–solvent pair. For an energetic model $\chi =A/T$, it occurs at $T=\Theta =2A$. More generally, if $\chi =A/T+B$, it occurs when $A/\Theta +B=1/2$. It is not the absence of all intermolecular interactions and does not require the polymer and solvent to be chemically identical.

Chain dimensions. Interpret $R$ as the root-mean-square end-to-end distance of a polymer chain. Let $a$ denote the supplied statistical step length, calibrated to the same count $N$ as the degree of polymerization. At the theta condition, random-walk statistics giveFor a good solvent, the Flory approximation balances entropic stretching and two-segment interaction:Minimization gives $R^5\sim v_{\mathrm{eff}}{a}^2{N}^3$, henceThus the familiar good-solvent scaling is $R\sim a{N}^{3/5}$ when its nonuniversal solvent-quality amplitude is suppressed. With the additional lattice estimate $v_0\simeq a^3$, the more explicit form is $R\sim a(1-2\chi {)}^{1/5}{N}^{3/5}$. This swollen-coil estimate is used away from the narrow theta crossover; it must not be extrapolated to predict $R=0$ as $\chi \uparrow 1/2$. The crossover occurs when $(1-2\chi )N^{1/2}$ is of order one. Nearer theta, the ideal-coil size supplies the appropriate limit.

Numerical sizes. The theta value isFor $\chi =0.3$, the plain good-solvent scaling givesIf the solvent-quality amplitude is retained with $v_0=a^3$, it instead givesThe distinction is an explicitly stated modeling convention: the question does not give a microscopic site volume or an absolute swollen-coil prefactor. Both estimates describe a coil larger than the theta coil. Here $(1-2\chi )\sqrt{N}\simeq 15.5$, so a swollen-coil estimate is reasonable. Neither result is an exact microscopic prediction from the supplied data alone.

Overlap concentration. The polymer overlap concentration corresponds to one chain per coil volume. Using the scaling convention $V_{\mathrm{coil}}\sim R_{\theta }^3$ givesThe first is chain number density; the second counts repeat units. Thus the repeat-unit molar concentration is approximately $0.143\mathrm{mol}{\mathrm{L}}^{-1}$. If the lattice-site volume is taken as $v_0=a^3$, this is equivalentlyFor a mass-concentration estimate, a polystyrene repeat unit ${\mathrm{C}}_8{\mathrm{H}}_8$ has molar mass about $104.15\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, so this scaling convention givesThere is no sharp coil surface fixing an exact numerical prefactor. For example, using the often adopted geometric definition $V_{\mathrm{coil}}=4\pi R_g^3/3$, with the ideal chain radius of gyration $R_g=R_{\theta }/\sqrt{6}=10.6\mathrm{nm}$, gives $c^{\ast }\simeq 52\mathrm{g}{\mathrm{L}}^{-1}$. The invariant content is one chain per coil volume and the theta scaling ${\varphi }^{\ast }\propto N^{-1/2}$; the radius and coil-volume convention must accompany a numerical value.

Osmotic pressure. In the given lattice expression, $\varphi$ is a dimensionless polymer volume fraction, not a dimensional number density, and $f$ is the mixing free energy per lattice site. Introduce $v_0$ to obtain a physical osmotic pressure. If $M=N_s+N{N}_p$ is the number of sites and $\varphi =N{N}_p/M$, differentiation of $F_{\mathrm{mix}}=Mf(\varphi )$ with respect to the solvent count at fixed chain count givesBalancing this change in solvent chemical potential against the pressure contribution $v_0\Pi$ gives $\Pi =(\varphi f^{\prime }-f)/v_0$. For composition-independent $\chi$,and therefore the osmotic pressure in Flory-Huggins solution theory isExpanding the logarithm as $\varphi \to 0$ givesThe first term is $n_p{k}_{B}T$ with $n_p=\varphi /(N{v}_0)$ now denoting chain number density: a whole chain supplies one translational contribution. The sign of the quadratic correction reproduces the solvent-quality classification. It is positive for a good solvent, zero at the theta temperature, and negative for a poor solvent. This is the same two-segment interaction that swells or contracts a single coil. At theta, a positive third-order contribution remains; the chain is approximately ideal in its large-scale conformation, not a gas of segments with no interactions at all.

First, planar Brownian motion started away from zero almost surely never hits zero. To see this without confusing it with neighborhood recurrence, stop in an annulus with inner radius $ϵ$ and fixed outer radius $R>|Z_0|$. The preceding logarithmic calculation gives probability $\log (R/|Z_0|)/\log (R/ϵ)$ of reaching the inner circle first. A hit of zero before the outer exit would require all these inner hits, so its probability is zero as $ϵ\downarrow 0$. On every finite path interval the continuous path is bounded; a countable union over integer outer radii and integer times proves the all-time assertion.

Consequently the integrand $1/Z$ is locally square integrable: on every compact time interval the nonvanishing continuous path has a positive minimum modulus. Write $Z=X+iY$ with independent standard real coordinates. Complex quadratic covariation satisfies $d[Z,Z]=dt-dt+2id[X,Y]=0$, so analytic Itô's lemma has no second-derivative drift. For $L=\int Z^{-1}dZ$ one likewise has $d[L,L]=0$, and $d[e^{-L},Z]=0$. The product rule therefore givesIts initial value is $Z_0$, proving, simultaneously for all times by continuity,Thus $\Re L_t=\log (|Z_t|/|Z_0|)$ and $\Im L_t$ is the continuous accumulated change of complex argument of the path about zero, measured in radians. It is a winding angle, without reduction modulo $2\pi$. A single-valued logarithm on the whole punctured plane is unnecessary; the path has a continuous stochastic logarithm of planar Brownian motion.

At $\alpha =0$ the transverse coordinates are independent Brownian motions with diffusion coefficient $D=1$. The cylinder is infinitely long, so the axial coordinate cannot affect its first contact time. By rotational invariance of Brownian motion, the mean first-passage time $u$ depends only on the transverse radius $\rho =\sqrt{x^2+y^2}$.

The Kolmogorov backward equation for a mean first-passage time is $Lu=-1$. Here $L={\partial }_x^2+{\partial }_y^2$, so the radial boundary value problem isThe absorbing boundary condition for a diffusion is imposed on the circular cross-section. At $\rho =0$, the solution must extend smoothly over the axis. Multiplying by $\rho$ and integrating givesSmoothness at the axis excludes the logarithm term, and $u(R)=0$ fixes $C_0=R^2/4$. ConsequentlyThis also verifies the finiteness assumption behind the Kolmogorov backward equation. Let $T$ be the first contact time and apply the stopped Itô formula to the bounded nonnegative function $u$ just obtained. Its bounded derivatives on the disk make the stopped stochastic integral a martingale, givingBy monotone convergence, $T$ has finite expected value; continuity and dominated convergence then make the final term vanish as $t\to \infty$. Hence $u({\hat{r}}_0)=\mathbb{E}T$. In particular, the mean first-passage time is $R^2/4$ on the axis and tends to zero at the adsorbing wall.

For a radial function $f(r)$ in three dimensions, the Laplacian is $f^{\prime \prime }+2f^{\prime }/r$, while the imposed radial drift contributes $-\alpha f^{\prime }/r$. Thus the radial generator of an SDE diffusion isThe hitting probability for the inner sphere is harmonic for this diffusion generator in the shell, with value one on the inner sphere and zero on the outer sphere. At $\alpha =1$, the boundary value problem is thereforeIntegrating $(r{p}^{\prime }{)}^{\prime }=0$ gives $p(r)=A\log r+B$. The two boundary conditions determineBoth arguments of the logarithms are dimensionless. The result decreases from one to zero across the shell and lies strictly between zero and one for $R_1<r_0<R_2$.

For a direct probability interpretation, stop at the first contact $T$ with either sphere. The stopped process $p(r(t\wedge T))$ is a bounded martingale by Itô's formula. The shell has positive inner radius, so its drift is bounded and its diffusion is nondegenerate; exit occurs almost surely. Optional stopping and dominated convergence therefore giveThe logarithm appears because the inward drift cancels one of the two geometric $f^{\prime }/r$ terms in the three-dimensional radial Laplacian, leaving the same radial diffusion generator as a planar Brownian motion with diffusion coefficient one.

Ensemble and applications. A grand canonical ensemble describes a system exchanging both energy and particles with a large reservoir. Its fixed control parameters are $T,V,\mu$, while microstates with different $N$ have probabilitiesWeak coupling allows the system's energy levels to remain meaningful; equilibrium fixes the reservoir temperature and chemical potential. The grand potential is $\Phi =-k_{B}T\ln \Xi =U-TS-\mu N$, with $N$ now the mean population. It generates entropy, pressure and mean particle number by differentiation. This ensemble is convenient for gases in contact with particle reservoirs, independently occupied quantum levels and surface adsorption; its particle-number fluctuations are part of the description, not an error in counting.

Classical single-level limit. For the single-level grand partition function in the dilute limit, put $x_k=e^{-\beta ({ϵ}_k-\mu )}$. Empty and singly occupied levels contribute $1$ and $x_k$. In the dilute classical limit $x_k\ll 1$, higher occupancies contribute only at order $x_k^2$ or beyond:This truncation does not impose the Pauli principle on a classical gas. A fermion state has exactly $1+x_k$, a boson state has $(1-x_k{)}^{-1}$, and Maxwell-Boltzmann occupancy weights give $e^{x_k}$; they all agree to this order. In particular, $\ln {\Xi }_k=x_k+O(x_k^2)$, with the classical Maxwell-Boltzmann result exactly $x_k$.

grand potential and chemical potential. With $z=e^{\beta \mu }$, summing the single-level logarithms and using the stated density of states givesSubstitute $ϵ=k_{B}T{x}^2$, for whichCombining this with the density-of-states prefactor gives $\ln \Xi =zV/{\lambda }^3$. ThusDifferentiating at fixed $T,V$,Similarly $p=-({\partial }_V\Phi {)}_{T,\mu }=k_{B}Tz/{\lambda }^3$, so $pV=N{k}_{B}T$. This is the grand-canonical chemical potential of a classical ideal gas. The condition $N{\lambda }^3/V\ll 1$ states the classical regime. The supplied density of states counts one internal state; extra internal degeneracies would have to be included consistently rather than inserted at only one stage.

One adsorption site. Take the free-vapour energy reference as zero and the binding energy as $-ϵ$. A site is either empty or singly occupied; the latter has vibrational partition function $z_s(T)$. Its adsorption-site grand partition function and mean occupation areThe chemical potential of an adsorbed species on equivalent sites follows by inverting the occupation ratioThe factor $1-\theta$ is the vacant-site probability. The binding reference and the energies included in $z_s$ must be chosen consistently, so that vibrational zero-point energy is not counted twice.

Isotherm. The vibrational factor in a Langmuir equilibrium constant is fixed by adsorption equilibrium, ${\mu }_s={\mu }_{\mathrm{vap}}$. Neglecting vapour internal degrees of freedom and writing its number density as $p/(k_{B}T)$ gives $e^{\beta {\mu }_{\mathrm{vap}}}=p{\lambda }^3/(k_{B}T)$. ThereforeThis is the Langmuir isotherm: initially linear, increasing monotonically and concave down, approaching unit coverage. Half coverage occurs at $p_{1/2}=1/b_s(T)$. The coefficient has inverse-pressure units. It assumes equivalent independent sites, at most one molecule per site and no lateral interaction or multilayer adsorption. The molecular mass appropriate to the vapour determines $\lambda$.

A real uptake measurement must separate adsorbed molecules from free gas in the apparatus. Langmuir's original mica experiment compared a mica-containing bulb with a similar blank and inferred uptake from calibrated gas volumes and pressures. His stated blank contribution and bath-temperature drift limited the inferred adsorption; a saturating curve alone does not establish ideal equivalent sites.^[ref]

Classical translational statistics. Write the number density as $n=N/V$ and the thermal de Broglie wavelength as $\lambda =\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$. Quantum exchange is negligible when wave packets rarely overlap:The equivalent state-counting statement is that the mean occupation of each thermally relevant one-particle state is much smaller than one. The equations in this question count one internal species; with $g$ equally populated spin states the relevant density per species is $n/g$.

Let $z=e^{\beta \mu }$ be the dimensionless quantum gas fugacity, with the lowest one-particle energy chosen as zero. Both quantum distributions can be writtenIf $z\ll 1$, then $n_{\eta }(E)\simeq z{e}^{-\beta E}\ll 1$ for all $E\ge 0$, recovering Maxwell–Boltzmann statistics. Conversely, low occupation even at the spectrum's bottom requires $z/(1-z)\ll 1$ for bosons or $z/(1+z)\ll 1$ for fermions; either condition impliesA single high-energy state can be sparsely occupied even in a degenerate gas, so the condition must hold across the relevant spectrum, not just for one arbitrarily chosen energy. Also, classical translational statistics does not imply that all internal excitation energies or other quantum effects can be ignored.

Grand partition function, density and pressure. Noninteracting particles have independent occupations of each one-particle state. Put $x_{\mathbf{k}}=z{e}^{-\beta {\epsilon }_{\mathbf{k}}}$. Summing the allowed occupations givesThereforeThe grand potential is ${\Omega }_G=-k_{B}T\ln \Xi =-pV$ for this homogeneous gas. Replacing the sum by $V\int d^{3}k/(2\pi {)}^3$ and using ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ givesThe mean number is $N=z{\partial }_z\ln \Xi$, givingFor bosons, the grand sum requires $z<1$ when the ground energy is zero. A macroscopic condensate at the limiting $z\to 1$ must be separated from the continuum integral in the condensed phase. The following small-$z$ calculation is in the normal, nondegenerate regime, so that complication does not enter.

Density expansion and inversion. Expand the occupation as a geometric series:The Gaussian wavevector integral for its $\ell$th term isIt follows thatFor the inverse series, put $y=n{\lambda }^3$ and try $z=y+c{y}^2+O(y^3)$. Substitution gives $0=(c+\eta /2^{3/2})y^2+O(y^3)$, henceThe leading equality $n{\lambda }^3\simeq z$ directly connects the density and occupancy criteria for classical behavior.

Pressure and its first quantum correction. Expanding the logarithm in the pressure expression gives one extra factor $1/\ell$ compared with the density series:Insert the inverse series. To second order in $y=n{\lambda }^3$,ThusThe first quantum correction at fixed density and temperature isIt is negative for bosons and positive for fermions, and vanishes in the classical density limit.

Virial coefficient. Define the number-density virial expansion byThe coefficient $B_j$ has units ${\mathrm{volume}}^{j-1}$ in this convention. Comparing terms givesThis is the exchange contribution to the second virial coefficient. It arises without a pair-interaction potential; the underlying gas is noninteracting. A molar-density convention would multiply this $B_2$ by the Avogadro constant. For $g$ equally populated internal species, exchange occurs within each species and the coefficient in terms of total density is smaller by a factor $g$.

Why the two signs differ. Bosonic exchange symmetry enhances the probability of nearby identical bosons, whereas fermionic exchange symmetry suppresses nearby identical fermions in the same internal state. In the dilute regime, the normalized equal-time pair detection correlation isThe exchange part follows by squaring the Gaussian one-particle coherence factor $e^{-\pi r^2/{\lambda }^2}$. At zero separation this gives $g^{(2)}(0)=2$ for bosons and $0$ for one-component fermions. Integrating the exchange pair weight provides a direct coefficient check:Bunching therefore acts like an effective statistical attraction in the dilute equation of state, lowering the pressure needed at fixed density. Exclusion acts like an effective statistical repulsion, raising it. Neither description asserts an actual interparticle force. Different internal fermion species do not exclude each other from a common spatial location in the same way, explaining the importance of the species convention.

The logarithm requires $y>0$, and $x>0$ is given. Set $z=\log (y/x)$, so $y=x{e}^z$ and the chain rule yields $x{y}^{\prime }/y=1+x{z}^{\prime }$. Dividing the ordinary differential equation by $y$ therefore givesSince $(z/x{)}^{\prime }=(x{z}^{\prime }-z)/x^2=0$, we obtain $z=Cx$, including the case $z=0$. Thus the general positive solution is $y=x{e}^{Cx}$. The initial condition $y(1)=e$ gives $C=1$:For the separate decay requirement, $x{e}^{Cx}\to 0$ precisely when $C<0$: if $C=0$, the result is $x$, and if $C>0$, it grows. Consequently the full family with the stated limit isThis latter family does not also satisfy $y(1)=e$; the solution selected by that initial condition grows. The identically zero function is excluded by the original logarithm.

For $r>0$, the chain rule gives $r_x=x/r$ and $r_y=y/r$. Applying it twice to the radial function $u=f(r)$ givesUsing $x^2+y^2=r^2$, Laplace's equation becomesTwo integrations now giveThese are the general solutions of the specified radial function form on an annulus or a punctured plane. If the solution must extend continuously to the origin, the logarithm term forces $A=0$, leaving only a constant function.

The logarithm laws giveFor $n\ge 2$, both $n\le n{\log }_{2}n$ and ${\log }_{2}n\le n{\log }_{2}n$. The expression is therefore at most $(3+{\log }_{2}5)n{\log }_{2}n$. ConsequentlyThe Big O notation concerns eventual behaviour; the zero value of $n{\log }_{2}n$ at $n=1$ is irrelevant.

The logarithm identity $a^{{\log }_{b}n}=n^{{\log }_{b}a}$ rewrites the expression as $n^{{\log }_{2}3}$. Since $3<4=2^2$, we have ${\log }_{2}3<2$, and for $n\ge 1$,Thus the Big O notation bound holds with $C=1$:

Repeated squaring computes $a,a^2,a^4,a^8,\dots$ by successively squaring. The binary representation of the exponent specifies which of these values to multiply. For modular exponentiation, reduce after every multiplication; only a number of steps proportional to the logarithm of the exponent is needed.

Here $13\equiv 2(\mathrm{mod}11)$ and $17=16+1$. Successive squares giveTherefore

For sufficiently large positive $n$, the logarithm identities giveAfter division by $n{\log }_{2}n$, the right-hand side tends to one. Thus $\boxed{\Theta (n\log n),\text{in particular }O(n\log n)}$ gives the asymptotic order in Big theta notation and Big O notation.

The logarithm is strictly increasing, so $\mathbb{P}(Y\le y)=\mathbb{P}(X\le e^y)$. Since $X>1$ almost surely,Thus the logarithm of a Pareto random variable has an exponential distribution with rate $c$. Its probability density function value at the single endpoint can be chosen arbitrarily without changing the law.

For natural logarithms and sufficiently large $n$, take the logarithm of the positive expression:Set $t=\log \log n$. Then $\log n=e^t$. The exponential function grows faster than $t^2$: for example, its power series gives $e^t\ge t^3/6$ for $t>0$, so $t^2/e^t\le 6/t\to 0$. In particular, eventually $(\log \log n{)}^2\le \log n$. Exponentiating yieldsFor logarithms to a fixed base $b>1$, the natural logarithm of the left side is $(\log \log n-\log \log b{)}^2/\log b$, still smaller than $\log n$ eventually. The big O notation conclusion is therefore independent of that base choice.

True. The logarithm identity givesFor $n\ge 1$, $e^n\ge n$, so $0\le n{e}^{-n}\le 1$. By monotonicity of the natural logarithm, the expression lies between $0$ and $\ln 2$. It is therefore $O(1)$ in Big O notation.

Use $f(x)=x{\log }_{2}x-4$. Then $f(1)=-4$, $f(2)=-2$, and the secant method givesWriting $L={\log }_{2}3$, the next step isThus the requested approximation is $x_3=2.73$ to three significant figures. All intermediate logarithms should be retained at higher precision before this final rounding.

True for every fixed $k\ge 0$. Extract $e^n$ from inside the logarithm:Choose an integer $m>k$. The nonnegative terms of the exponential series give $e^n\ge n^m/m!$ for $n>0$, whenceSince $0\le \log (1+u)\le u$ for $u\ge 0$, the displayed expression tends to zero and is therefore $O(1)$. The constant and cutoff may depend on the fixed $k$; no uniform assertion over unbounded $k$ is being made.

Eliminate the constraint by writing $y=1/x$, with $x>0$. The logarithms cancel, so the objective becomes $q(x)=1/x+2x$. Its derivatives are $q^{\prime }(x)=-1/x^2+2$ and $q^{\prime \prime }(x)=2/x^3>0$. Thus the unique stationary point is the unique global minimum, at $x=1/\sqrt{2}$. Restoring $y$ givesAs a direct global check, the arithmetic-geometric mean inequality gives $1/x+2x\ge 2\sqrt{2}$, with equality exactly at the same point.

Integrate the solids’ molar volume contribution from $P^{\circ }=1$ bar and use ideal-gas water chemical fugacity $f=P$. For pure solids,The logarithm is dimensionless, and $\Delta V_s=-2.28\mathrm{J}\mathrm{ba}{\mathrm{r}}^{-1}$ can be multiplied directly by pressure in bars. Therefore the dehydration equilibrium with ideal-gas fluid givesHence $T=1295.57$ K, or $1022.42^{\circ }$C. Do not also add a constant gas-volume pressure term: the water contribution has already been integrated through $RT\ln (P/P^{\circ })$. Ideal-gas behavior at $6$ kbar is a prescribed model approximation, not a realistic equation of state. The solid volume correction is distinct from it.