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For discrete random variables, write $p(u,v)$ for the joint probability mass function and $p_U(u)$ for its marginal. The conditional entropy is the average of the entropies of the conditional distributions:Conditional distributions at zero-probability values of $u$ can be chosen arbitrarily; they contribute nothing. Here $h$ denotes Shannon entropy, with one common logarithm base and $0\log 0=0$.

For $p(u,v)>0$, use $\log p(u,v)=\log p_U(u)+\log p(v\mid u)$ and sum with weight $-p(u,v)$. Summing out $v$ in the first term givesFor countable alphabets, the entropy contributions are nonnegative, so the equality remains valid with extended values by summing nonnegative terms. Applying this identity successively to the prefix $(X_1,\dots ,X_{i-1})$ and $X_i$ proves the chain rule for information entropy:For $i=1$, the conditioning tuple is empty and the term is $h(X_1)$.

For the subset inequalities, first suppose the marginal entropies are finite, so all joint entropies below are finite by the chain rule and the assumed conditioning reduces entropy property. Put $H=h(X_1,\dots ,X_n)$, and let $H_{-i}$ be the joint entropy with coordinate $i$ omitted. Apply the two-variable chain rule in the order $(X_{[n]\setminus \{i\}},X_i)$ to getThe inequality holds because the prefix conditioning set is contained in the set of all other coordinates. Summing over $i$ and using the chain rule givesThis proves Han's entropy inequality. In terms of the normalized subset entropy, $h_n^{(n)}=H/n$ and $h_{n-1}^{(n)}={\sum }_i{H}_{-i}/[n(n-1)]$, hence $\boxed{h_n^{(n)}\le h_{n-1}^{(n)}}$ for $n\ge 2$.

Apply the same result to each $k$-element coordinate set $S$. It givesWhen this is summed over all $k$-subsets, each fixed $(k-1)$-subset $T$ occurs once for each possible added coordinate, namely $n-k+1$ times. ThereforeSubstitute ${\sum }_{|S|=k}h(X_S)=k\left(\genfrac{}{}{0ex}{}{n}{k}\right)h_k^{(n)}$ and the analogous identity at $k-1$. The binomial coefficient identity $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=(n-k+1)\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)$ cancels the common positive factor and givesThis is the monotonicity of normalized subset entropies, obtained from the local $k$-coordinate inequality by explicit counting.

If some marginal entropy is infinite, each subset average contains at least one subset including that coordinate. The entropy of such a subset is at least that marginal entropy, by the nonnegative conditional-entropy chain rule, so every $h_k^{(n)}$ is infinite. The inequalities then hold in the extended sense. Otherwise the finite-entropy proof applies. For independent random variables the averages are all $n^{-1}{\sum }_{i}h(X_i)$; repeated copies of one variable instead give $h_k^{(n)}=h(X_1)/k$, illustrating the entropy reduction caused by redundancy.

Let $Y$ be the number of isolated vertices and set $\lambda =e^{-c}$. For fixed $k\ge 1$, the falling factorial $(Y{)}_k$ counts ordered $k$-tuples of distinct isolated vertices. A specified tuple is isolated exactly when all edges incident to its vertices are absent. There are $k(n-k)+\left(\genfrac{}{}{0ex}{}{k}{2}\right)$ such edges, soSince $np=\log n+c$ and $n{p}^2=o(1)$, taking logarithms givesThus every fixed factorial moment tends to ${\lambda }^k$. We use the standard factorial-moment criterion for Poisson convergence: nonnegative integer-valued random variables whose factorial moments converge, for every fixed order, to ${\lambda }^k$ converge in distribution to $\mathrm{Poisson}(\lambda )$. Consequently $Y$ has that limiting distribution, and its mass at each fixed integer converges. In particular,This establishes the limiting probability of at least one isolated vertex, rather than merely a bound based on its expectation.

The Poisson distribution mass function isIt is an exponential dispersion family of order one withIndeed, $b^{\prime }(\theta )=e^{\theta }=E(Y)$ and $\varphi b^{\prime \prime }(\theta )=e^{\theta }=\mathrm{Var}(Y)$. A generalized linear model specifies independent random variables from a response exponential family, together with a linear predictor and a link function relating it to the response mean. The Poisson canonical link is $g(\mu )=\log \mu$.

For the insurance counts, let class $1$ and merit $0$ be the reference levels in a regression factor. The fitted Poisson regression assumesThe insured exposure $n_{ij}$ is known; its logarithm is an offset, not a coefficient to estimate. The model has eight free mean parameters and fixes the dispersion parameter at one. It assumes additive class and merit effects on the log-rate scale, hence multiplicative effects on the rate scale without a class-by-merit interaction.

Apart from constants independent of the parameters, the log-likelihood isDifferentiating gives the Poisson regression margin-matching score equationsThus the fitted totals equal the observed totals in each class and merit category, including the reference categories by subtraction. In design matrix notation these equations are $Z^T(Y-\hat{\mu })=0$. Fisher scoring solves them iteratively; the reported three iterations describe the numerical fit, not an additional statistical result.

The regression intercept gives the baseline fitted claim rate $e^{-2.0357359}=0.130584$ per insured car year. The merit rate ratios, relative to merit $0$ at the same class, areThe class rate ratios, relative to class $1$ at the same merit, areEach predicted cell rate is the baseline rate times its class and merit multipliers. The output's coefficient-to-standard error ratios are asymptotic normal Wald test statistics under the fitted Poisson regression, despite the software label “t value”; there is no estimated Gaussian residual scale in this model. All nonreference effects have very large absolute ratios.

The null Poisson deviance is $33854.16$ on nineteen statistical degrees of freedom; adding the seven factor coefficients reduces it by $33274.6437$. But the final Poisson deviance is still $579.5163$ on twelve statistical degrees of freedom, and the deviance residuals include values near $-10.79$ and $11.63$. These are far beyond what a well-fitting unit-dispersion model would normally produce. The additive Poisson model fits substantially better than a common rate, but remains grossly inadequate. A class-by-merit interaction, overdispersion, or statistical dependence between claims could contribute; the summary alone does not distinguish these explanations. Its very small nominal standard errors and formal Wald tests rely on the rejected model, so improved mean structure and error assumptions are needed before treating them as reliable uncertainty assessments.

For $|\alpha |>1$, an exterior polynomial root of $\rho$ persists by continuity for sufficiently small negative real $z$ in the stability polynomial $\rho (w)-z\sigma (w)$. Hence these parameters cannot be A-stable.

For $-1\le \alpha <1$, the principal polynomial root at $w=1,z=0$ is simple. Let $s(z)=\log w(z)$ denote its analytic logarithm near zero. Using the expansion in part (b), and ${\rho }^{\prime }(1)=2(1-\alpha )$, solve the polynomial root equation to obtainIndeed the residual at $s=z$ is $-(\alpha +5)z^4/12$, and the correction in $s$ cancels it by multiplication with $2(1-\alpha )$. Choose $z=-C{\epsilon }^4/2+i\epsilon$. This lies strictly in the left half-plane, butfor small positive $\epsilon$. Thus $|w(z)|=e^{\mathrm{Re}s(z)}>1$, proving linear instability within the A-stability domain. This is a direct principal-root obstruction to A-stability, rather than an appeal to an order-barrier theorem alone.

Finally, at $\alpha =1$, the full stability polynomial isIt retains a double unit polynomial root for every negative $z$, allowing growing parasitic solutions. ThereforeThe canceled Backward Euler method recurrence at $\alpha =1$ is A-stable, but its admissible solutions and starting relations differ from those of the original method.

For each fixed $x>0$, continuity of the logarithm gives the pointwise convergence $f_n(x)\to \log x$. But at $x_n=1/n$, which belongs to $(0,1)$ for $n>1$,Consequently the uniform error cannot tend to zero. In fact $f_n(x)-\log x=\log (1+1/(nx))$ is unbounded as $x\downarrow 0$ for every fixed $n$. The pointwise limit is $\log x$, and convergence is not uniform.

Let $A(x)$ count primes $p\le x$ for which $p+2$ is also prime. We prove $A(x)\ll x/{\log }^{2}x$ using a quadratic Selberg sieve, then apply Abel summation.

For $F(n)=n(n+2)$, each odd prime has exactly two roots modulo that prime. The Chinese remainder theorem gives the polynomial root density in a sievefor odd squarefree integers $d$. Put $R=x^{1/8}$ and choose the optimizing real Selberg sieve weights ${\lambda }_d$, supported on these $d\le R$, with ${\lambda }_1=1$. If $p,p+2$ are prime and $p>R$, their weight $({\sum }_{d\mid F(p)}{\lambda }_d{)}^2$ is one. It is nonnegative for every other integer. Expanding the square therefore givesThe Selberg sieve diagonalization makes the main quadratic form $1/J(R)$, whereThe optimal Selberg weights have modulus at most one. Since ${\rho }_F([d,e])\le [d,e]\le R^2$, the remainder is at most $O(R^4)$.

For completeness, the lower bound on $J$ is where the second logarithm enters. Restrict to primes $3\le p\le y=R^{1/8}$ and put $k(p)=2/(p-2)$. The full product $W={\prod }_{3\le p\le y}(1+k(p))$ is $\asymp {\log }^{2}y$, since its logarithm is $2{\sum }_{3\le p\le y}1/p+O(1)$ and the Mertens second theorem applies. Give the associated squarefree divisor $d$ probability $k(d)/W$. Prime inclusion probabilities are $2/p$, so the Mertens first theorem givesfor large $R$. The Markov inequality shows that at least half the weight has $d\le R$. This proves the truncated Euler-product lower bound $J(R)\gg W\gg {\log }^{2}R$. Consequently the twin-prime upper bound from a quadratic sieve isFinally, Abel summation givesThe tail is dominated by ${\int }_3^{\infty }dt/(t{\log }^{2}t)<\infty$. Thus reciprocal-sum convergence from a counting bound proves convergence of the first reciprocal sum. The second is smaller termwise. We have proved Brun's theorem:Counting each twin prime just once changes at most a finite repetition and gives the same conclusion.

Use base-two logarithms, so Von Neumann entropy is measured in bits, and set $0\log 0=0$. Write $D(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$ for the quantum relative entropy, the quantity denoted $S(\rho \Vert \sigma )$ in the paper. It is $+\infty$ unless the support of a positive operator $\rho$ lies within that of $\sigma$.

We first establish nonnegativity of quantum relative entropy, including its equality condition. Let $\rho ={\sum }_{i:r_i>0}{r}_i|u_i⟩⟨u_i|$ and $\sigma ={\sum }_j{s}_j|v_j⟩⟨v_j|$, and suppose the support condition holds. Put $q_i=⟨u_i|\sigma |u_i⟩>0$. Concavity of the scalar logarithm givesTerms with $s_j=0$ have zero weight here. Working first with natural logarithms, the inequality $-\ln x\ge 1-x$ yieldssince the $q_i$ sum to at most $\mathrm{Tr}\sigma =1$. Equality forces $q_i=r_i$ for every positive-weight eigenvector, no remaining weight of $\sigma$ outside their span, and equality in the logarithm's strict concavity. The latter makes each $u_i$ an eigenvector of $\sigma$ with eigenvalue $q_i$. Thus $D(\rho \Vert \sigma )=0$ exactly when $\rho =\sigma$.

For the bipartite density operator,Indeed, a projector onto the kernel of ${\rho }_A$, tensored with $I_B$, has zero expectation in ${\rho }_{AB}$. Positivity implies that ${\rho }_{AB}$ annihilates its range: $\mathrm{Tr}({\rho }_{AB}P)=\Vert {\rho }_{AB}^{1/2}P{\Vert }_{\mathrm{HS}}^2=0$. The same argument applies to $B$. Hence the quantum relative entropy below is finite. On this support,Using the partial trace to evaluate the two local terms givesThis proves Subadditivity of Von Neumann entropy:The proof also identifies equality exactly with the product state ${\rho }_{AB}={\rho }_A\otimes {\rho }_B$.

A nondegenerate distribution function $G$ is a max-stable distribution if, for every integer $r\ge 1$, there are $a_r>0$ and $b_r\in \mathbb{R}$ such thatEquivalently, the suitably centered and positively scaled sample maximum of $r$ independent and identically distributed random variables with law $G$ again has law $G$. The extremal types theorem states that every nondegenerate limiting law of normalized sample maxima is a max-stable distribution. Thus this property identifies all possible limiting types, although a particular starting distribution function need not admit any such limiting normalization.

The three standard types are the Gumbel distributionthe Fréchet distribution, with $\alpha >0$,and the negative Weibull distribution, with $\alpha >0$,They are max-stable distributions, as follows respectively by taking $(a_r,b_r)=(1,\log r)$, $(r^{1/\alpha },0)$, and $(r^{-1/\alpha },0)$. The necessary and sufficient classification is: $G$ is max-stable exactly when $G(x)=H((x-b)/a)$ for $a>0$, $b\in \mathbb{R}$, and one of these three types $H$. A positive affine map sends a max-stable distribution to another max-stable distribution.

For the requested threshold equivalence, set $p_n=1-F(u_n)$. By independence,If $n{p}_n\to \tau <\infty$, then $p_n\to 0$ and the expansion $\log (1-p)=-p+O(p^2)$ gives $n\log (1-p_n)\to -\tau$, since $n{p}_n^2=(n{p}_n)p_n\to 0$. Exponentiation proves the claimed limit. Conversely, if $(1-p_n{)}^n\to e^{-\tau }>0$, then $p_n\to 0$: any subsequence with $p_n\ge \delta >0$ would have its powers tending to zero. Taking logarithms now gives $-n\log (1-p_n)\to \tau$. Since $-\log (1-p)/p\to 1$, including the continuous extension at $p=0$, it follows that $n{p}_n\to \tau$. Hence

The exceedance count $S_n$ has the binomial distribution with parameters $n,p_n$. For $\tau >0$ and each fixed nonnegative integer $s$, its probability mass function satisfieswhere $(n{)}_s=n(n-1)\cdots (n-s+1)$ and $(n{)}_0=1$. Indeed $(n{)}_s/n^s\to 1$, $(n{p}_n{)}^s\to {\tau }^s$, and $(1-p_n{)}^{n-s}\to e^{-\tau }$. If $\tau =0$, the bound $\mathbb{P}(S_n\ge 1)\le \mathbb{E}{S}_n=n{p}_n\to 0$ gives the degenerate Poisson distribution at zero. Summing the finitely many terms proves, in both cases,This derives the needed instance of the Poisson limit theorem directly.

Finally fix $x$ with $G(x)>0$ and put $u_n=a_{n}x+b_n$. The assumed limit for the normalized sample maximum and the proved equivalence give $n{p}_n\to -\log G(x)$. The second-largest order statistic is at most $u_n$ exactly when at most one observation is strictly greater than $u_n$. This event identity holds even if observations have ties. Apply the preceding Poisson limit theorem with $k=1$ to obtainThe endpoint $G(x)=1$ corresponds to $\tau =0$ and is already included. This is the case $r=2$ of the exceedance count limit for extreme order statistics.

For a pro-p group, the lower p-series is the descending sequence of closed characteristic subgroupsThe bar means closure of the subgroup generated by the displayed powers and group commutators. Throughout this question $p$ is odd. For a finitely generated powerful pro-p group, group commutator collection gives $P_i(G)=G^{p^{i-1}}$, each $P_i$ is powerful, and $\Phi (P_i)=P_{i+1}$. All these subgroups are open. A uniform pro-p group is such a group with all subgroup indices $[P_i:P_{i+1}]$ equal to $[G:P_2]$. We use the standard powerful-group torsion criterion: a finitely generated powerful pro-p group is uniform exactly when it is a torsion-free group. For a uniform group of generator number $d$, the indices are $p^d$, and $x\mapsto x^{p^a}$ is a homeomorphism from $G$ onto $G^{p^a}$.

We record explicitly the collection facts used for the operations below. With $U_b=G^{p^b}=P_{b+1}$, one hasThe second statement is the power-congruence property: the $p$th-power map is an isomorphism on adjacent uniform layers, and collection gives $(yu{)}^p\equiv y^p{u}^p(\mathrm{mod}{U}_{b+2})$ for $u\in U_b$. Thus the first nonzero layer of a difference moves down exactly one step under a $p$th power. Iterating gives the asserted equivalence and also justifies taking roots of congruences. These facts follow from powerful-group group commutator collection, using divisibility by $p$ of $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ and the successive-layer power isomorphisms; they are the uniform-group structure results being used here.

The transported addition on a uniform pro-p group isThe product is in $U_n$ and has a unique $p^n$th root in $G$. For the comparison at $n=1$ define $x{+}_{0}y=xy$. The numbered proofs below establish associativity, approximate commutativity and compatibility of these operations.

The intrinsic Lie algebra of a uniform pro-p group has underlying set $G$, zero equal to the group identity, additive inverse $x^{-1}$, and operationsHere the group commutator is $[x,y]=x^{-1}{y}^{-1}xy$. For $\lambda \in {\mathbb{Z}}_p$, $x^{\lambda }$ is the limit of $x^{m_j}$ for integers $m_j\to \lambda$; the pro-$p$ topology makes it well defined. The addition sequence is Cauchy by the compatibility proved below. The group commutator estimate puts $[x^{p^n},y^{p^n}]$ in $U_{2n+1}$, so its displayed root exists and lies in $U_1$. Collection of group commutators shows that successive bracket approximants differ in layers tending to the identity, giving the other limit. The group commutator identities give additivity of the limiting bracket, and the Hall-Witt identity gives its Jacobi identity. Approximate commutativity makes limiting addition commutative. The resulting lattice $L_G$ is a free module over ${\mathbb{Z}}_p$ of rank $d(G)$, and $[L_G,L_G]\le p{L}_G$.

A powerful Lie algebra over p-adic integers, in the convention relevant here, is a Lie algebra $L$ whose underlying module is finite free over ${\mathbb{Z}}_p$ and whose bracket satisfies $[L,L]\le pL$. For $p=2$ the condition is instead $[L,L]\le 4L$. For odd $p$, define multiplication on its underlying set by the convergent Baker--Campbell--Hausdorff formulaAn iterated bracket of degree $m$ lies in $p^{m-1}L$. The BCH coefficient bound loses at most $\lfloor (m-1)/(p-1)\rfloor$ powers of $p$, so all higher-degree terms are integral, and their valuations tend to infinity. This explains both convergence and why multiplying a lattice with small bracket stays inside the lattice. Formal associativity of BCH passes to the convergent series, the identity is $0$, the inverse is $-x$, and every group power satisfies $x^{\ast m}=mx$. The quotients by $p^{a}L$ are finite p-groups. Hence the resulting compact group is a finitely generated torsion-free group and a pro-p group, and its group commutator lies in $pL$. Thus $(L,\ast )$ is a uniform pro-p group, with $P_i(L,\ast )=p^{i-1}L$ and successive index $p^{{\mathrm{rank}}_{{\mathbb{Z}}_p}L}$.

The standard uniform-group/Lie-lattice correspondence identifies these constructions as mutually inverse: $L_{(L,\ast )}\cong L$ and $(L_G,\ast )\cong G$. To see why the inverse limits recover the lattice operations, the addition approximant for a BCH group is $p^{-n}\mathrm{BCH}(p^{n}x,p^{n}y)$: its linear term is $x+y$ and every higher term tends to zero. Similarly its group commutator approximant has limiting term $[x,y]$. In the other direction, a local logarithm in the complete normed group algebra identifies the intrinsic operations with algebra addition and commutator and identifies group multiplication with BCH; the uniform power maps extend the identification from a sufficiently small power subgroup to all of $G$. Continuous group homomorphisms respect powers, unique roots and limits, while Lie algebra homomorphisms respect every BCH term, so the equivalence also preserves morphisms.

The objective is a continuous faithful representation of a uniform pro-p group $G$ in some ${\mathrm{GL}}_N({\mathbb{Z}}_p)$. The proof proceeds from the intrinsic Lie lattice to a small matrix group, then extends the resulting representation to all of $G$. This last step matters: a representation defined only on an open subgroup does not yet establish linearity of the original group.

Let $L=L_G$ be the Lie algebra of a uniform pro-p group and put $\mathfrak{g}={\mathbb{Q}}_p{\otimes }_{{\mathbb{Z}}_p}L$. The uniform-group/Lie-lattice correspondence identifies the group law on $L$ with BCH. The substantial finite-dimensional Lie-theoretic input is the Ado theorem: every finite-dimensional Lie algebra over a characteristic-zero field admits a faithful finite-dimensional representation. Apply it to obtainAn Adjoint representation would not suffice here, since it kills the centre; Ado supplies injectivity even in central directions.

Choose a p-adic lattice $\Lambda$ spanning $V$. Since $L$ has finitely many lattice generators, their representing matrices have bounded denominators. There is therefore an integer $a\ge 0$ such thatOn this matrix lattice the series $\exp X={\sum }_{n\ge 0}{X}^n/n!$ and $\log (1+Y)={\sum }_{n\ge 1}(-1{)}^{n+1}{Y}^n/n$ converge and are inverse. For example $v_p(n!)\le n/(p-1)$, so the valuations of the exponential terms tend to infinity with this choice of $ϵ$; the logarithm terms do as well. They take values in $1+p^{ϵ}\mathrm{End}(\Lambda )$ and in $p^{ϵ}\mathrm{End}(\Lambda )$ respectively. The formal identity $\exp (\mathrm{BCH}(X,Y))=\exp X\exp Y$ holds for the convergent series.

Under the correspondence, $U=G^{p^a}$ has Lie lattice $p^{a}L$ and is an open characteristic subgroup. Definewhere ${\log }_G$ denotes the identification of $U$ with $p^{a}L$. A Lie homomorphism preserves BCH, so the exponential identity proves $\eta (uv)=\eta (u)\eta (v)$. If $\eta (u)=I$, the matrix logarithm gives $\rho ({\log }_{G}u)=0$; faithfulness of $\rho$ gives $u=1$. Thus $\eta$ is a continuous faithful integral representation of $U$.

To extend it, take the moduleValues on a transversal of $U\backslash G$ determine $f$, so $W$ is a free module over ${\mathbb{Z}}_p$ of rank $[G:U]\dim V$. These functions are continuous, since on each open coset they are given by the continuous representation $\eta$. Define $(T_{g}f)(x)=f(xg)$. The equivariance condition is preserved, and $T_g{T}_h=T_{gh}$. In a transversal basis each $T_g$ is a permutation of blocks followed by matrices from $\eta (U)$, so it belongs to $\mathrm{GL}(W)$ and depends continuously on $g$.

Faithfulness can be checked directly. If $g\notin U$, right multiplication moves the coset $U$ to a different coset, so $T_g$ moves a function supported on one block and cannot be the identity. If $g\in U$ and $T_g=1$, evaluation at $1$ gives $f(g)=f(1)$ for every $f$, whereas equivariance gives $f(g)=\eta (g)f(1)$. Since $f(1)$ is arbitrary, $\eta (g)=I$, and hence $g=1$. ThereforeCompactness of $G$ makes this continuous injection a homeomorphism onto its closed image. This proves the linearity of a uniform pro-p group for the full group, including its centre, and explains why taking a sufficiently small power lattice permits matrix integration before finite-index induction restores the whole group.

The Prime number theorem says $\pi (x)\sim x/\log x$ as $x\to \infty$. Bertrand's postulate says that for every integer $n\ge 2$ there is a prime number strictly between $n$ and $2n$.

To count S-smooth numbers, split each prime exponent into its parity and an even part. This gives the unique expression $n=a{b}^2$, where $a$ is a squarefree product of a subset of $S$. There are $2^{|S|}$ possible $a$, while $b\le \sqrt{x}$ whenever $n\le x$. The squarefree-part bound for smooth numbers therefore gives $f_S(x)\le 2^{|S|}\lfloor \sqrt{x}\rfloor \le 2^{|S|}\sqrt{x}$.

For an integer $x\ge 1$, take $S$ to be all prime numbers at most $x$. Every positive integer at most $x$ is then S-smooth, so $x\le 2^{\pi (x)}\sqrt{x}$. Taking logarithms yields

Use base-two logarithms, so entanglement entropy is measured in bits. Write $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$ for the binary entropy, with $0{\log }_{2}0=0$. Taking a partial trace over any two parties removes the off-diagonal terms: the two traced strings are orthogonal. Consequently all three one-party reduced density matrices of the first pure state areEach division into two groups is one party versus the other two. The Schmidt coefficients are $\sqrt{p}$ and $\sqrt{1-p}$, so the entanglement entropy across each of $A:BC$, $B:AC$ and $C:AB$ is $h_2(p)$. For the second pure state, set $p=1/2$: every one-party reduced density matrix is $I/2$, and all three entanglement entropies are one bit.

For every pair $XY$, the first pure state instead gives the reduced density matrixThis is a separable quantum state, explicitly a convex combination of two product states. Its relative entropy of entanglement is therefore zero: in $E_R(\rho )={\inf }_{\sigma \in \mathrm{Sep}}D(\rho \Vert \sigma )$, choose $\sigma =\rho$ to attain zero, while nonnegativity of quantum relative entropy excludes a negative value. The same argument with equal weights applies to the second pure state. ThusThe pairwise density matrices contain classical correlations, despite the entanglement of the global pure state; discarding a party loses the coherence between its two branches.

The entanglement entropy of $n$ independent source copies across any cut is $n{h}_2(p)$, whereas $f(n)$ copies of the target GHZ state have entanglement entropy $f(n)$. Average monotonicity of pure-state entanglement entropy under LOCC gives $P(n)f(n)\le n{h}_2(p)$ for exact successful forward branches. In a reversible asymptotic conversion the reverse direction similarly prevents a first-order deficit, since $n^{\prime }/n\to 1$. For asymptotically faithful outputs, a continuity bound for quantum conditional entropy with trivial conditioning changes these inequalities only by $o(n)$. Hence the reversible leading-order rate isThe zero pairwise relative entropies of entanglement are consistent with this conversion: no independent pairwise resource has to be supplied or removed.

There is a qualification to the literal finite-block formulation in the source. For $0<p<1$, the Schmidt rank of the exact state of $n^{\prime }$ source copies is $2^{n^{\prime }}$, whereas $f$ target GHZ states have Schmidt rank $2^f$. Every refined LOCC branch acts by a product of local linear operators and cannot increase this Schmidt rank. For $p\ne 1/2$, the displayed rate has $f<n^{\prime }$ for sufficiently large blocks when $n^{\prime }/n\to 1$, so an exact successful reverse conversion is impossible, irrespective of its claimed probability. The exact dilution obstruction from Schmidt rank is avoided in the usual asymptotic formulation by allowing outputs approaching the desired pure state. The leading-order rate above and the exact forward protocol below remain valid; neither establishes the source's stronger exact round-trip claim. At $p=1/2$ the two states are identical. At $p=0$ or $1$, the source is a known product state, the rate is zero, and local preparation supplies any desired number of source copies.

Divide metal enrichment by gas depletion in the closed-box model of galactic chemical evolution:The logarithm appears on integrating from $M_0$ and $Z=0$:With the gas fraction of a galaxy $\mu =M_g/M_0$, this isIt gives $Z=0$ at $\mu =1$ and increasing gas-phase metallicity as gas is consumed, independently of the time dependence of the star formation rate. For this closed initially star-free system, $M_0=M_g+M_s$, so this initial-reservoir fraction equals the present gas-to-total baryonic fraction. That equality fails in an open system.

First, for $0<x<1$, the derivative of $h(x)=\log (1-x)+x$ is $-x/(1-x)<0$, while $h(0)=0$. Hence $\log (1-x)<-x$.

Choose $N>c$ large enough that the given ratio bound holds for every $n\ge N$. Positivity permits taking logarithms, and the preceding inequality givesSumming, and using the integral test for convergence from part (b), gives for $n>N$Thus $a_n\le a_N{N}^c{n}^{-c}$. Since $c>1$, the P-series $\sum n^{-c}$ converges, so the comparison test for series provesThe finitely many initial terms do not affect convergence. This power-law comparison from a refined ratio bound supplies information when the ordinary ratio test can be inconclusive.

Put $V=|v_0|>0$, take $q>0$ without loss of generality, and write $s=R^2+z^2/q^2$. The Newtonian gravitational potential is singular at the origin; all pointwise claims below concern $s>0$. The zero of the Newtonian gravitational potential is the one in the paper, with an implicit fixed length unit inside the logarithm.

The Poisson equation for Newtonian gravity in cylindrical coordinates givesThe two contributions to the Laplacian areConsequently the mass density of the axisymmetric logarithmic gravitational potential isThe coefficient of $R^2$ is positive. The coefficient of $z^2$ determines the density positivity for an axisymmetric logarithmic potential: strict positivity at every noncentral point requires $q^2>1/2$ and $v_0\ne 0$. Nonnegativity permits $q^2=1/2$, where the mass density vanishes on the noncentral symmetry axis. For $q^2<1/2$ it is negative near that axis. Either sign of a nonzero real $q$ gives the same model, and $q=0$ is undefined. Setting $v_0=0$ yields vacuum away from the singular origin, not strictly positive mass density. The $r^{-2}$ central cusp is locally integrable and contains no hidden point mass: the Newtonian gravitational field flux through a shrinking sphere is of order its radius.