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Every allocation is determined by the set of $n$ children on the red bus, so there are $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ equally likely allocations. Exactly one places every boy on red and every girl on green. HenceApplying Stirling formula to the numerator and denominator gives the central binomial coefficient asymptoticand therefore

The Von Mangoldt divisor identity givesBy the Stirling formula, the left side is $\log (\lfloor x\rfloor !)=x\log x-x+O(\log x)$, while the assumed Chebyshev estimate gives $\psi (x)=O(x)$. HenceThe contribution of proper prime powers is bounded uniformly:Removing it leaves

For one molecule whose translational motion separates from its internal degrees of freedom, the classical one-particle partition function iswhere $q_{\mathrm{int}}$ contains rotational, vibrational, and other internal states. For $N$ indistinguishable noninteracting molecules, the classical ideal-gas partition function is $Z_N=z_1^N/N!$. The thermodynamic pressure therefore obeyssoThe internal structure changes the energy, entropy, and heat capacity through $q_{\mathrm{int}}(T)$, but it does not change the ideal gas law while the internal partition function is independent of volume and the molecules have negligible interactions.

For a dilute monatomic gas $q_{\mathrm{int}}=1$. Using the Stirling formula in $Z_N=(V/{\lambda }^3{)}^N/N!$ and differentiating the Helmholtz free energy $F=-k_{B}T\log Z_N$ gives the Sackur-Tetrode equationWith $N$ fixed, an isentropic process therefore satisfies $V{T}^{3/2}=\text{constant}$, or $T{V}^{2/3}=\text{constant}$. Eliminating $T$ with the ideal-gas law yields the monatomic reversible ideal-gas adiabat

Let two systems exchange energy inside an otherwise isolated composite system, so $E_1+E_2=E$ is fixed. By the Second law of thermodynamics, equilibrium maximizes the total entropyAt an interior maximum,Thus $T_1=T_2$, which is entropy maximization under thermal contact and the condition for thermal equilibrium.

Stability requires this entropy maximum to be locally strict, so each ordinary subsystem has a concave entropy-energy relation. Sinceone has $\partial T/\partial E>0$, or equivalentlyThis is the entropy concavity and positive heat capacity criterion.

Now let $N_{\uparrow }=\alpha N$ and $N_{\downarrow }=(1-\alpha )N$. Because the spins are distinguishable, the number of microstates is the binomial coefficientThe Boltzmann entropy $S=k_B\log \Omega$ and the Stirling formula give, to leading order for large $N$,The total energy isThereforeSolving for the up-spin fraction gives the independent spin-one-half two-level system resultFor ordinary positive absolute temperature,with $\alpha \to 0$ as $T\to 0^{+}$ and $\alpha \to 1/2$ as $T\to +\infty$. Since the spectrum is bounded above, the population-inverted range $1/2<\alpha \le 1$ is also mathematically possible and corresponds to negative temperature; $\alpha =1/2$ is the infinite-temperature state.

Substitution givesand direct differentiation yieldsThus the energy increases with temperature throughout either finite-temperature branch, in particular throughout the required positive-temperature range.

For the simple random walk on the integer line, a return to the origin is possible only at an even time. At time $2n$, exactly $n$ of the increments must be $+1$, so the return probability isThe supplied factorial bounds, equivalently the order estimate in the Stirling formula, giveConsequently ${\sum }_n{p}_{2n}$ diverges. By the recurrence criterion by return probabilities, the origin is a recurrent state; translation invariance then makes the whole walk recurrent.

For three independent walks, the probability that all three are at the origin at time $2n$ isThis P-series is convergent, so the first of the Borel-Cantelli lemmas says that simultaneous returns occur only finitely often with probability one. Therefore the requested probability is

Begin with the Von Mangoldt divisor identitySumming it for $n\le x$ and reversing the order givesThe given bound $\psi (x)\ll x$ and the Stirling formula therefore imply

Apply partial summation with the weight $1/\log n$. Writing $A(t)=\log t+E(t)$, where $E(t)=O(1)$, givesIndeed, the integral of $E(t)/(t(\log t{)}^2)$ converges, and its tail from $x$ to infinity is $O(1/\log x)$.

Grouping the left side by prime powers yieldsThe full double series over $k\ge 2$ converges. Its tail beyond $x$ is $O(x^{-1/2})$: split at $p=\sqrt{x}$, use a geometric series for $p\le \sqrt{x}$, and compare ${\sum }_{p>\sqrt{x}}{p}^{-2}$ with the corresponding sum over integers. Absorbing its limit into the constant proves the Mertens theorem for reciprocal primes

Write the second form of the functional equation asFor $-2\le \sigma \le 2$, the elementary exponential formula for the sine gives, uniformly for $t\ge 4$,The stated Stirling formula givesuniformly on the same strip. The bounded factors $2^{\sigma }{\pi }^{\sigma -1}$ and the cancelling exponentials therefore show thatTaking absolute values in the functional equation proves the Vertical-strip factor in the Riemann zeta functional equation:

For $S_n\sim \mathrm{Pois}(n)$,ConsequentlyPart c says that this tends to $1/\sqrt{2\pi }$. Rearranging gives the Stirling formula

Write the summand as $a_n=x^n/[\Gamma (n)n^n]$. By Stirling formula,SetThen the exponential phase isIt has a unique maximum at $t=1$, where $f(1)=2$ and $f^{\prime \prime }(1)=-2$. The contributing indices satisfy $n-N=O(\sqrt{N})$, so their width tends to infinity and the lattice sum may be replaced by a Riemann sum. The Discrete Laplace method therefore givesHence

The mean extension in an ensemble of $A$ molecules iswith constraints ${\sum }_i{a}_i=A$ and ${\sum }_i{a}_i{l}_i=AL$. For occupancies $\{a_i\}$, the number of assignments of the $A$ distinguishable ensemble members to the states is the multinomial coefficientThus maximizing the probability at fixed $A$ and $L$ is equivalent to maximizing the stated Lagrange multiplier expression.

Because $A\gg 1$, the Stirling formula gives $\log (a_i!)\simeq a_i\log a_i-a_i$. Differentiating with respect to each $a_i$ givesNormalization by ${\sum }_i{a}_i=A$ therefore yields the probability mass functionHere $Z$ is the fixed-tension partition function that normalizes the probabilities.

Only even times contribute, and at time $2k$ the walk is at zero exactly when it has made $k$ steps in each direction. By linearity of expectation,The Stirling formula implies thatHence there is a constant $c_0>0$ such that this term is at least $c_0/\sqrt{k}$ for every $k\ge 1$, after decreasing $c_0$ to cover the finitely many small values. ThereforeThe final expression is $2c_0(\sqrt{n+1}-1)$, so, after another adjustment for small $n$, there is a constant $c>0$ withfor every $n$. This is the lower bound recorded by expected visits to the origin by a simple random walk.

For $k\ge 0$, take $f_k(x)=1_{\{x\ge k+1\}}$ and write $q_k=\mathbb{P}(X\ge k+1)$. Since $f_k^2=f_k$,Its discrete derivative is nonzero only at $k$, soFor a Poisson distribution, $p_{k+1}/p_k=\lambda /(k+1)$ and $q_k\sim p_{k+1}$ as $k\to \infty$. Hencewhere the final estimate follows from Stirling formula. No finite constant $C$ can therefore make the proposed inequality hold for every $f$.

The Borsuk conjecture asserted that every bounded subset of ${\mathbb{R}}^d$ of positive diameter can be partitioned into $d+1$ subsets of strictly smaller diameter. We construct a Kahn-Kalai counterexample to the Borsuk conjecture.

For every $2p$-subset $A$ of $[4p]$, let $v_A\in \{-1,1{\}}^{4p}$ be $1$ on $A$ and $-1$ outside it, and defineBecause $x_A=x_{A^c}$, retain one representative of each complementary pair. The resulting set $X$ has $\frac{1}{2}\left(\genfrac{}{}{0ex}{}{4p}{2p}\right)$ points. For $s=|A\cap B|$,andAll $x_A$ have the same norm, so their distance is largest exactly when this inner product is smallest, namely when $s=p$.

Every smaller-diameter part of $X$ therefore corresponds to a family with no pair having intersection $p$. Part ii bounds such a part by $2\left(\genfrac{}{}{0ex}{}{4p}{p-1}\right)$. Any smaller-diameter partition consequently needs at leastparts. By Stirling formula, this ratio grows like $(27/16{)}^p$ up to a polynomial factor, whereas $d+1=O(p^2)$. For every sufficiently large prime $p$, the required number of parts exceeds $d+1$, disproving the conjecture.

In Laplace method for the Gamma function, scaling $t=x(1+s/\sqrt{x})$ gives a standard Gaussian integral. The first even correction polynomial is $s^2-7s^4/12+s^6/18$. Its expectation under a unit-variance centered Gaussian is $1-7\cdot 3/12+15/18=1/12$. The amplitude expansion contributes to this coefficient as well as the phase expansion; expanding only the latter gives the wrong Stirling formula correction.

For $X_j$ the number of tree components of order $j$ in a binomial random graph, choose their vertices, choose one of $j^{j-2}$ labelled trees by the Cayley formula, require its $j-1$ edges, and exclude both the remaining internal edges and all crossing edges. For $j=1$ interpret $j^{j-2}=1$, recovering the isolated vertex count. If $p=\lambda /n$ for fixed positive $\lambda$ and $j=O(\log n)$, the Stirling formula gives $\mathbb{E}{X}_j\sim n(\lambda e^{1-\lambda }{)}^j/(\lambda \sqrt{2\pi }{j}^{5/2})$. Distinct overlapping vertex sets cannot both be graph components; for disjoint sets their joint occurrence gains the factor $(1-p{)}^{-j^2}$ relative to the product, because their between-set edges must be absent only once.

Writingthe Stirling formula gives, uniformly on every fixed vertical strip and for large $|t|$,