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For $N$ distinguishable noninteracting spins, let a fraction $\alpha$ occupy the level $+ϵ$ and the remainder occupy $-ϵ$. The Stirling formula givesThe relation $1/T=\partial S/\partial E$ yieldsPositive temperature corresponds to $0\le \alpha <1/2$; a population inversion $1/2<\alpha \le 1$ has negative temperature.

For $\psi (t)={\sum }_{m\le t}\Lambda (m)$, finite rearrangement and the Von Mangoldt divisor identity giveThe Stirling formula, or integral comparison for the logarithm, gives $T(x)=x\log x-x+O(\log (x+1))$ for real $x\ge 1$.

Let $\mu$ be the Möbius function. Its Möbius divisor-sum identity isIndeed, for $n>1$ the sum is the expansion of ${\prod }_{p\mid n}(1-1)$; nonsquarefree divisors contribute zero. Thus, for arithmetic functions $a,b$,To prove Möbius inversion, substitute the first formula into the second and collect the coefficient of $a(e)$. It is ${\sum }_{d\mid n/e}\mu (d)$, which is one for $e=n$ and zero otherwise. Conversely the same divisor interchange recovers $b$ from the displayed formula for $a$. In Dirichlet convolution notation this is simply $\mu \ast 1=\epsilon$, where $\epsilon$ is supported at one.

The Prime number theorem with classical zero-free-region error states that some absolute $c>0$ satisfiesHere $\Lambda$ is the Von Mangoldt function. An equivalent prime-counting form, after decreasing $c$ if needed, is $\pi (x)=\mathrm{Li}(x)+O(x{e}^{-c\sqrt{\log x}})$. The contributions of proper prime powers are $O(\sqrt{x}{\log }^{2}x)$ and can be absorbed into this error. We use $\Psi$ to keep it distinct from the Fourier transform appearing in Question 5.

Here is a direct deduction of the requested Mertens bound from a log-integrable Chebyshev error from the stated Prime number theorem. First, the divisor identity givesSecond, the exact Dirichlet convolution identityfollows from $\log =1\ast \Lambda$: multiplication of a Dirichlet convolution by $\log n$ differentiates its two factors, so applying it to $\mu \ast 1=\epsilon$ and convolving again with $\mu$ gives this formula. Summing it yieldsThe last harmonic sum is $O(1)$. To see this without an endpoint approximation, split $x/d$ into dyadic ranges $[2^j,2^{j+1})$. In each range $\sum 1/d=O(1)$, and its exponential factor is at most $e^{-c\sqrt{j\log 2}}$. Their sum over $j\ge 0$ converges. The finitely many terms with $1\le x/d<2$ obey the same estimate after adjusting the constant in the prime-number-theorem error. ThereforeFor the Mertens function $M(x)={\sum }_{n\le x}\mu (n)$,The second sum has absolute value at most ${\sum }_{n\le x}\log (x/n)=O(x)$, by an integral comparison or the Stirling formula. ConsequentlyThe absolute value in this conclusion is present in the original PDF.

Assume $0<\lambda =1-\epsilon <1$, as required for this nontrivial subcritical binomial random graph. Put $\delta =\lambda -1-\log \lambda >0$. A breadth-first exploration of a binomial random graph is dominated by a Galton-Watson process whose offspring counts are independent random variables with law $\mathrm{Bin}(n,\lambda /n)$. If its total progeny $T$ is at least $j$, the first $j-1$ offspring counts sum to at least $j-1$. Their moment-generating function is bounded by that of a Poisson distribution of mean $\lambda (j-1)$. The exponential Markov inequality, optimized at $t=-\log \lambda$, yieldsTaking $j=\lceil (1+\eta )\log n/\delta \rceil$ for any fixed $\eta >0$, a union bound over all starting vertices gives $\mathbb{P}(L_1\ge j)=o(1)$.

For the matching lower bound, let $X_j$ count tree components of order $j$. The tree-component expectation in the Erdős-Rényi model is exactlyFor $j=O(\log n)$, the Stirling formula givesChoose $j=\lfloor (1-\eta )\log n/\delta \rfloor$, with $0<\eta <1$. Then $\mathbb{E}{X}_j\to \infty$. Distinct overlapping vertex sets cannot both be graph components. For disjoint sets, the joint probability differs from the product only because the $j^2$ between-set edges were counted twice as absent. ThusIt follows that $\mathrm{var}{X}_j/(\mathbb{E}{X}_j{)}^2=o(1)$. The second moment method gives $X_j>0$ with high probability. Combining both bounds, and then taking arbitrarily small fixed $\eta$, proves the sharp subcritical largest-component scaleFinally the Taylor expansion of $-\log (1-\epsilon )$ gives $\delta ={\epsilon }^2/2+{\epsilon }^3/3+\cdots$. The endpoint $\lambda =0$ has no edges and is excluded from this logarithmic asymptotic.

Watson's lemma applies to Laplace-type endpoint integrals such as ${\int }_0^b{e}^{-\lambda t}f(t)dt$ as $\lambda \to +\infty$, when $f$ has a local algebraic expansion $f(t)\sim {\sum }_j{c}_j{t}^{p+j-1}$ with $p>0$ and suitable integrability or exponential-growth control away from the endpoint. Each term contributes $c_j\Gamma (p+j){\lambda }^{-p-j}$. The exponential selects a neighborhood of $t=0$.

Laplace method instead treats ${\int }_a^{b}g(t)e^{\lambda \varphi (t)}dt$ by locating the dominant maxima of a real phase $\varphi$. A smooth nondegenerate interior maximum produces a neighborhood governed by a Gaussian function of width ${\lambda }^{-1/2}$; endpoint or degenerate maxima require their corresponding local scaling. Contributions away from a strict dominant maximum must be smaller. These methods overlap after suitable changes of variable, but their usual starting forms emphasize an endpoint amplitude expansion and an exponential maximum, respectively.

For the Gamma function, put $t=xu$. ThenThe phase has its unique maximum at $u=1$, with value $-1$ and second derivative $-1$. Set $u=1+s/\sqrt{x}$. Near this maximum,Combining the amplitude and exponential expansions in Laplace method givesLocalization near $u=1$ justifies extending the limits for the Gaussian integral; the distant endpoint is not being expanded uniformly by this local series. Odd terms integrate to zero. The moments of the Gaussian integral satisfy $I_1=I_0$, $I_2=3I_0$, $I_3=15I_0$, so the relative correction isThe two terms of the Stirling formula, obtained by the Stirling correction by Gaussian moments, are therefore

A Markov chain is irreducible if every state can reach every other state in finitely many steps with positive probability. It is a recurrent Markov chain if, starting at any state, it returns to that state with probability one; for an irreducible chain this property is shared by all states.

For the simple symmetric random walk on $\mathbb{Z}$, taking $|j-i|$ steps all in the appropriate direction reaches $j$ from $i$ with probability $2^{-|j-i|}>0$. Thus the chain is irreducible. Its return probabilities areBy Stirling formula, $p_{00}^{(2k)}\sim 1/\sqrt{\pi k}$, so their sum diverges. The standard recurrence criterion ${\sum }_{n\ge 0}{p}_{00}^{(n)}=\infty$ therefore proves the walk is recurrent.

The printed hint's constant is erroneous: the actual limit of $\sqrt{k}{p}_{00}^{(2k)}$ is $1/\sqrt{\pi }$. Its positivity is what the recurrence argument requires, but the reciprocal is necessary for the correct numerical value.

In a point-particle description of quantum gravity, interactions of a massless spin-two field occur at localized vertices of Feynman diagrams. The gravitational coupling has negative mass dimension for spacetime dimension greater than two, so short-distance loop integrations generate higher-derivative counterterms that cannot all be absorbed into the Einstein action. A string worldsheet instead joins incoming and outgoing strings by a smooth surface. In particular, a closed string splits or joins through a smooth pair-of-pants surface. There is no invariantly distinguished point at which every part of an extended string interacts. The choice of a time slicing can draw a joining point, but it is not a physical pointlike vertex. This removes the point-interaction geometry responsible for arbitrarily localized interactions and introduces the string length ${\ell }_s=\sqrt{{\alpha }^{\prime }}$.

This geometric explanation is supported by the organization of the Polyakov path integral. Introducing a worldsheet metric makes the Nambu–Goto action amenable to gauge fixing. In the Euclidean integral, metrics on a fixed oriented surface are divided by worldsheet diffeomorphisms and Weyl transformations. What remains is a Riemann surface with a conformal structure, together with finitely many worldsheet moduli and the positions of external string vertex operators. The determinants from fixing the metric are represented by Faddeev-Popov ghosts. Schematically, for $n$ closed-string external states,Here ${\mathcal{M}}_{g,n}$ is the moduli space of genus-$g$ punctured surfaces. The dilaton weights a surface by its Euler characteristic $\chi =2-2g$, giving $g_s^{-\chi }$ before the normalization of external states. Every additional handle multiplies the contribution by $g_s^2$. The string genus expansion therefore plays the role of the ordinary loop expansion, with the sphere giving tree level and the torus giving one loop. The critical dimension cancels the worldsheet Weyl anomaly, making this reduction to conformal geometry consistent.

The massless symmetric traceless closed-string state is a graviton, and its long-wavelength interactions reproduce gravity. At distances comparable with ${\ell }_s$, however, an infinite tower of excited states contributes. This is more than a single particle form factor: the same amplitude contains graviton exchange, massive higher-spin exchange and string-scale softness.

The dimensionless variables in the given Virasoro–Shapiro amplitude correspond to $s={\alpha }^{\prime }S/4$, $t={\alpha }^{\prime }T/4$, $u={\alpha }^{\prime }U/4$, where capital letters denote the usual dimensionful Mandelstam invariants. Its constraint $s+t+u=-4$ identifies the external particles as the closed bosonic tachyons, not four gravitons. Nevertheless its intermediate massless pole directly exhibits graviton exchange. The overall normalization will be suppressed below; no additional copy of the amplitude formula is needed.

First, the amplitude is symmetric under permutations of $s,t,u$, expressing crossing symmetry. The poles in any channel areFor generic $t$, the residue can be calculated with the Gamma function pole at $-n$ and its recurrence:The product is $1$ when $n=0$. Indeed, the pole factor contributes $(-1{)}^{n+1}/n!$, while the two remaining Gamma ratios contribute $(-1{)}^n(t+2{)}_n^2/n!$. A residue polynomial of degree $2n$ shows exchange of spins up to $2n$, with leading Regge trajectory $J=2+2s$. At $n=1$, the massless residue is $-(t+2{)}^2$, containing the spin-two exchange. By crossing, the massless $t$-channel pole isfor generic $s$. Thus at large $s$ the pole behaves as $s^2/t$, the characteristic energy dependence of graviton exchange. Its massless propagator produces the long-range gravitational interaction. At low energies the massive poles can instead be expanded into local higher-derivative corrections to the effective gravitational action.

At the string scale one must retain the whole tower. The Gamma reflection formula rewrites the amplitude exactly asAt large $s$ and fixed $t$, away from the resonance poles or with a specified complex continuation, the Gamma ratio scales as $s^{t+1}$. Hence the Regge limit scales as $s^{2+2t}$, up to the signature factor and the displayed $t$-dependent coefficient. In particular, a fixed nonzero momentum transfer changes the point-graviton power law by a factor $s^{2t}$: the graviton belongs to an entire trajectory rather than remaining an isolated spin-two exchange at arbitrarily high energy.

There is stronger suppression when the scattering angle is fixed. Put $t\simeq -as$, $u\simeq -(1-a)s$ with $0<a<1$. Applying Stirling formula to the Gamma functions, after the same treatment of real-axis poles, gives the leading smooth envelopeThe leading $s\log s$ terms cancel and the remaining entropy-like combination is negative. In physical units the exponent is $-{\alpha }^{\prime }Sf(a)/2$. This fixed-angle softness of the closed-string amplitude is absent for an elementary pointlike gravitational vertex. Equivalently, short-distance gravitational scattering probes additional string degrees of freedom. In a four-dimensional long-distance effective setting, a massive exchanged mode contributes a Yukawa-type correction proportional to $e^{-M_{n}r}/r$, with couplings and tensor factors depending on the external states. Such terms are suppressed for $r\gg {\ell }_s$ and become relevant for $r\sim {\ell }_s$. The amplitude establishes the spectrum and short-distance change; it does not by itself give a universal static potential for arbitrary sources. Its bosonic tachyon also signals an unstable vacuum, so it must not be treated as a complete stable model of a gravitational force.

Finally, the modular group prevents overcounting equivalent surfaces and reorganizes the loop ultraviolet region. A torus is $\mathbb{C}/(\mathbb{Z}+\tau \mathbb{Z})$ with $\mathrm{Im}\tau >0$. A change of lattice basis identifiesThus the one-loop integral uses a fundamental domain for $\mathrm{PSL}(2,\mathbb{Z})$, for exampleThe measure $d^2\tau /(\mathrm{Im}\tau {)}^2$ is invariant, and the physical integrand, including matter, ghosts and external insertions, must respect torus modular invariance. In this domain $\mathrm{Im}\tau \ge \sqrt{3}/2$. The region of arbitrarily small proper time that causes ultraviolet divergences in a particle loop is therefore not an independent part of the string integral; a modular transformation relates it to another description already represented in the domain.

This does not remove every possible divergence. The cusp $\mathrm{Im}\tau \to \infty$ is a long tube, and factorization through it can produce infrared divergences, massless tadpoles or the bosonic tachyon divergence. At higher genus, degenerating cycles similarly factorize into propagating intermediate string states. The essential improvement is the relation between worldsheet geometry, the full spectrum and the quotient by large diffeomorphisms; infrared and unstable-background problems remain separate questions. A short-distance expansion that keeps only the graviton discards precisely this structure.

Let $S_0=0$ and let each increment of the simple symmetric random walk be uniformly chosen from $\{\pm e_1,\pm e_2,\pm e_3\}$. A return requires an even number of steps. In $2n$ steps returning to zero, let $a,b,c$ be the respective numbers of positive steps in the three directions; there must also be $a,b,c$ negative steps, so $a+b+c=n$. Counting step sequences givesThe terms $p_{abc}=n!/(3^{n}a!b!c!)$ form a multinomial distribution, so $\sum p_{abc}^2\le (\max p_{abc})\sum p_{abc}=\max p_{abc}$. The permitted combinatorial inequalities give $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\le C_1{n}^{-1/2}$ and ${\max }_{a+b+c=n}{p}_{abc}\le C_2/n$ for $n\ge 1$. The second bound also follows by noting that the largest multinomial coefficient occurs when $a,b,c$ differ by at most one and then applying Stirling formula. ThusTo prove the origin is a transient state, let $q$ be the probability of at least one return after time zero. The Strong Markov property at successive returns gives $\mathbb{P}(N\ge k)=q^k$ for the number $N$ of later returns. Therefore $\mathbb{E}(N+1)={\sum }_j{u}_j$ would be infinite if $q=1$. The bound forces $q<1$, proving that the origin is a transient state. Translation invariance gives the same conclusion at every lattice point. The coordinate walks need not be treated as independent; the counting via multinomial coefficients already incorporates their dependence.

For the one-dimensional freely jointed chain, counting right- and left-pointing links givesThereforeChoosing which links point right gives the state degeneracyPut $x=L/(Na)$. Applying the Stirling formula to $S=k_B\log \Omega$ and discarding terms subleading in $N$ yields

Stirling formula states