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For $t>0$, the natural logarithm satisfiesTo prove it, put $h(t)=\ln (1+t)-2t/(2+t)$. Then $h(0)=0$ and the derivative is $h^{\prime }(t)=t^2/[(1+t)(2+t{)}^2]>0$. More precisely,This positive integral controls errors in symmetric approximations to the logarithm. After writing $x=1+t$, the bound is $\ln x>2(x-1)/(x+1)$ for $x>1$; the inequality reverses for $0<x<1$ by replacing $x$ with $1/x$.

For $r\ge 0$, integration gives the survival function and cumulative distribution functionSolving $F_R(r_p)=p$ gives the quantile function^[ref]

For the chosen positioning model $\sigma =2\mathrm{cm}$,The disk of radius one component standard deviation therefore contains about 39.35% of outcomes; the familiar one-dimensional normal interval percentage does not apply to a two-dimensional disk. The 95% and 99% radial thresholds areA circle defined by $R=r_p$ itself still has probability zero; its enclosed disk carries the stated probability.

The closed-form quantile function also gives inverse transform sampling: if $U$ has uniform distribution on $(0,1)$, then $\sigma \sqrt{-2\log U}$ has Rayleigh distribution. Using $1-U$ instead gives the same law. Endpoints are excluded so that the logarithm stays finite.

Let $P$ be a joint probability distribution on a finite product alphabet, with marginals $P_i$. Compare it with a product model $Q={⨂}_i{Q}_i$, assuming $Q_i(x_i)>0$ wherever $P_i(x_i)>0$. Adding and subtracting the logarithm of ${\prod }_i{P}_i(x_i)$ givesThe first term is the unavoidable error from discarding dependence; the other terms are errors in the individual marginal probability distributions. Averaging a term that depends only on $x_i$ under $P$ is the same as averaging it under $P_i$, which proves the identity directly.

Among product laws, the choice $Q_i=P_i$ minimizes this relative entropy, leaving exactly total correlation. The related cross-entropy equals entropy plus relative entropy identity interprets this as extra average logarithmic loss from using the independent model. It is a statement about a specified joint law and loss function, not a claim that fitting a finite sample automatically recovers that law.

For $|r|<1$, let $S_N$ contain the first $N$ terms, with $S_0=0$. The remainder after those terms isThis is the exact absolute error. A convenient bound replaces the denominator by $1-|r|$, giving $|R_N|\le |a||r{|}^N/(1-|r|)$ through the triangle inequality. The exact denominator is preferable when the sign or complex phase of $r$ is known.

For $a>0$ and $0<r<1$, the omitted fraction of the entire sum is $R_N/(a/(1-r))=r^N$. Capturing at least $99$ percent therefore means $r^N\le 0.01$. Taking logarithms and applying the ceiling function gives the smallest integer $N$:This gives $7,21,90,$ and $459$ terms for $r=0.5,0.8,0.95,$ and $0.99$, respectively. The difference is a concrete reason to care about the common ratio, even when all four series converge.

An absolute target has a different scale. For $a=1$, $r=0.8$, and required absolute error at most $0.01$, the total sum is $5$ andThe first acceptable term count is $N=28$: the remainder is approximately $0.0096714$, while at $N=27$ it is approximately $0.0120893$. Counting terms and checking the tail avoids an ambiguous instruction to stop when a term merely looks small.

The replicator equation says that a strategy grows in frequency when its expected payoff exceeds the population's current mean payoff. It is an ordinary differential equation for a composition, rather than for total population size. The subtraction of the mean is what makes relative success matter: even a type growing in absolute number can shrink as a fraction if other types grow faster.

Let $x_i(t)$ be the frequency of pure strategy $i$, with $n$ available pure strategies andIn a symmetric bimatrix game, let the payoff matrix $A$ have entry $a_{ij}$ equal to the payoff to strategy $i$ against $j$. The other player's payoff in that encounter is $a_{ji}$: symmetry between player roles does not require $A=A^T$. Uniform random encounters give the expected payoffs and their population meanAfter choosing the time unit so that the growth-rate scale is one, the continuous-time replicator equation isHere a payoff is measured on a scale proportional to net growth per unit time, not merely as an arbitrary preference ranking. More generally one can specify locally Lipschitz frequency-dependent payoffs $f_i(x)$ directly; the same equation with $\bar{f}={\sum }_i{x}_i{f}_i$ then defines a unique frequency trajectory. The linear pairwise model and its reproductive interpretation are given in Taylor and Jonker (1978), sections 2–3; the formulas here also follow by the following direct normalization.

Suppose the number $N_i$ of type $i$ obeys ${\dot{N}}_i=N_i[b(t)+f_i(x)]$, where the baseline $b(t)$ is identical for all types. Set $N={\sum }_i{N}_i>0$ and $x_i=N_i/N$. The quotient rule givesThus the common baseline cancels. Adding the same constant to every entry of $A$ leaves the replicator equation unchanged; multiplying $A$ by a positive constant changes only the speed along trajectories. The same payoff matrix can therefore describe frequency evolution at different growth-rate baselines and time scales.

The probability simplex is invariant in forward time. On ${\sum }_i{x}_i=1$,Normalization alone would not exclude negative frequencies. Nonnegativity follows as well: along a solution,An absent type remains absent, and a positive frequency remains positive at every finite time. For the fixed payoff matrix model, bounded rates on the compact probability simplex give global forward solutions. Frequencies can still tend to zero as $t\to \infty$. Each face is invariant, so this model has no spontaneous introduction of new pure strategies; mutation or outside introductions would require additional terms. For two present types the logarithm of their frequency ratio satisfies $\frac{d}{dt}\log (x_i/x_j)=f_i-f_j$, directly expressing the accumulated payoff advantage.

For two pure strategies, write the composition as $(x,1-x)$, where $x$ is the frequency of strategy 1, and letTheir payoff difference and the resulting scalar ordinary differential equation areIn the interior $0<x<1$, the sign of the payoff difference determines the direction of motion. The endpoints are always equilibrium points, and an additional interior equilibrium point exists precisely when the affine payoff difference has a zero strictly between them. At a simple interior zero $x_{\ast }$, $F^{\prime }(x_{\ast })=x_{\ast }(1-x_{\ast })\Delta f^{\prime }(x_{\ast })$: a negative value gives asymptotic stability, a positive value gives instability. At the endpoints $F^{\prime }(0)=b-d$ and $F^{\prime }(1)=c-a$, with stability understood relative to $[0,1]$. A zero derivative requires examining the actual flow rather than this strict sign test.

For an original coordination game, imagine two conventions that reward matching, with convention 1 giving the larger matching reward. ChooseThe crossing is $x_{\ast }=2/5$. Below it, convention 1 loses frequency; above it, convention 1 gains frequency. The endpoint equilibrium points $0$ and $1$ have one-sided asymptotic stability, while $2/5$ is unstable: the corresponding derivatives are $-2,-3$ and $6/5$, respectively. Every initial $0<x(0)<2/5$ tends to $0$, every initial $2/5<x(0)<1$ tends to $1$, and exactly $x(0)=2/5$ stays there. Even the convention with the larger matching reward can disappear if it starts below the threshold. This is bistability, not selection of a universally best convention.

The compositions giving a symmetric Nash equilibrium in this coordination game are the two pure conventions and the mixed composition $(2/5,3/5)$. Both pure conventions are evolutionarily stable strategies because a mutant convention has strictly lower payoff against the pure resident. The mixed symmetric Nash equilibrium is not an evolutionarily stable strategy: against that resident both conventions tie, but after a pure convention-1 mutant is introduced its payoff advantage is $5(x-2/5)>0$. Equal payoffs at a symmetric Nash equilibrium need not provide resistance to invasion or asymptotic stability.

For an original coexistence model, choose payoffs making each convention advantageous when sufficiently rare:Here $x_{\ast }=3/4$ attracts every initial composition with both types present: $\dot{x}>0$ below it and $\dot{x}<0$ above it. The interior derivative is $F^{\prime }(3/4)=-3/4$, while $F^{\prime }(0)=3$ and $F^{\prime }(1)=1$ make both endpoints unstable to introduction of the other type. The endpoints nevertheless remain stationary if the other type is exactly absent; the attraction statement excludes those boundary initial states.

The sole symmetric Nash equilibrium in this coexistence model is $p=(3/4,1/4)$. It is also an evolutionarily stable strategy. For any different mixed strategy $q=(y,1-y)$, the first comparison in the two-condition criterion for evolutionary stability ties because $Ap=(3/2,3/2)$. Its second comparison is strictly positive:Thus restoring frequency feedback can protect a mixture even though all present types earn equal payoffs at the mixture. Coexistence here means different individuals using different pure strategies; it does not require every individual to randomize with a mixed strategy.

Frequency evolution need not increase the mean payoff. In the coexistence example $\bar{f}=5x-4x^2$, whose maximum occurs at $x=5/8$, whereas the attracting composition is $3/4$. For $5/8<x<3/4$, the frequency $x$ increases while $\frac{d}{dt}\bar{f}=(5-8x)\dot{x}<0$. Individual payoff advantages drive the equation; maximizing collective mean payoff is a different optimization problem.

More generally, a rest point only equalizes payoffs among present types: $x_i>0$ implies $(Ax{)}_i=x^{T}Ax$. A symmetric Nash equilibrium additionally requires $(Ax{)}_i\le x^{T}Ax$ for every absent type. All vertices are rest points even when an absent type could invade. For finite symmetric bimatrix games, a rest point with Lyapunov stability on the full probability simplex is a symmetric Nash equilibrium; an interior trajectory that converges has a symmetric Nash equilibrium as its limit. These are conditional results, not a convergence theorem for every trajectory. See Cressman and Tao (2014), Theorem 1.

The rock paper scissors payoff matrix gives an explicit obstruction to universal convergence:Since $A^T=-A$, $x^{T}Ax=0$, and in the interior $\frac{d}{dt}\log (x_1{x}_2{x}_3)={\sum }_i(Ax{)}_i=0$. Thus $x_1{x}_2{x}_3$ is conserved. Its maximum on the probability simplex occurs only at $(1/3,1/3,1/3)$, the unique symmetric Nash equilibrium, so an interior trajectory starting elsewhere cannot converge to that composition. Every other interior product level is a closed curve without a rest point, giving a periodic orbit. The existence of these cycles is also discussed in Cressman and Tao (2014), Example 1. The useful lesson is to inspect the specified payoff feedback and dynamical system, rather than infer convergence from the existence of an evolutionarily stable strategy or a Nash equilibrium.

Residual-offset uncertainty in nail thickness estimates is the uncertainty contributed by an imperfectly known additive background after correction of spectral signals. It differs from a known uncorrected offset's bias: subtraction can remove the best estimate of a background while leaving uncertainty about that estimate.

For an original reduced optical model, let two measured normalized signals be $Y_1,Y_2$ and subtract the same background estimate $B$. Define $Z_j=Y_j-B>0$ and suppose the attenuation contrast $D={\mu }_1-{\mu }_2\ne 0$ is known. The inferred nail plate thickness isIts sensitivity coefficients areWith mutually independent raw signal estimates and background estimate, first-order uncertainty propagation givesThis follows the sensitivity-and-covariance approach in the measurement guide.^[ref]

For stipulated $Z_1=0.30$, $Z_2=0.50$, $D=0.50\mathrm{m}{\mathrm{m}}^{-1}$, $u(Y_1)=u(Y_2)=0.003$ and $u(B)=0.010$, the estimated thickness is approximately 0.5108 mm and its combined standard uncertainty is 0.0177 mm. The common-background contribution alone is 0.0133 mm. Averaging 100 independent repetitions of each raw signal while reusing the same background estimate reduces the total only to about 0.0134 mm; it does not reduce that shared contribution by ten.

An equivalent calculation treats corrected signals as correlated: $\mathrm{Cov}(Z_1,Z_2)=u^2(B)$ under the stated independence assumptions. Omitting this covariance would describe a different experiment with independent background errors. Near zero corrected signal, the logarithm also becomes strongly nonlinear, so this local approximation needs further checking.

The primary nail experiment included a fluorescence correction in its optical transmission measurement.^[ref] Background emission and dark current must be characterized separately when they have different spectral responses. The equations above are an illustrative uncertainty budget, not a validated thickness calibration or reported precision from that experiment.

Suppose a calibrated measurement bounds the fractional loss of a nonzero superconducting persistent current by $ϵ$, with $0<ϵ<1$, over a duration $T>0$. Take $I_0>0$ by choosing the circulation direction. Under the constant-$L$, constant-$R$, undriven RL circuit model,The inequality follows because the logarithm is increasing. The final approximation requires $ϵ\ll 1$ and slightly understates the exact upper bound. If $L$ is uncertain, an independently established upper bound on $L$ gives the conservative electrical resistance bound. The uncertainty in the initial reading, the final reading, background subtraction and drift must all be included in $ϵ$; a display resolution alone does not provide it.

For an original illustrative example, let $L=1\mu \mathrm{H}$ and suppose the corrected electric current ratio after $T=3\mathrm{h}=10800\mathrm{s}$ is at least $0.999$. ThenThese are conditional bounds for stipulated parameters, not values attributed to the 1914 apparatus. With the same $L$ but $R=10^{-11}\Omega$, the model instead gives $\tau =10^5\mathrm{s}$ and $I(24\mathrm{h})/I_0=e^{-0.864}\simeq 0.42147$: a readily visible loss on the chosen scale.

A resolved decay can be checked using its logarithm:A straight line supports this constant-parameter model; curvature or jumps demand a different account. If an effective electrical resistance varies with time while $L$ remains fixed and the loop remains undriven, only its time average is constrained:This does not bound every instantaneous value of $R(t)$. Changing self-inductance or external linked magnetic flux adds terms to the circuit equation, so an uncorrected magnetic signal cannot simply be substituted for $I(t)$. Nonobservation of decay leaves arbitrarily small positive constant electrical resistance compatible with a finite observation: it is not a logical proof that $R=0$.

Relative to the specified solid, $Q_{\mathrm{sp}}<K_{\mathrm{sp}}$ favors dissolution if that solid is available; $Q_{\mathrm{sp}}>K_{\mathrm{sp}}$ favors precipitation; equality marks saturation. These comparisons identify a driving direction favored by thermodynamics, not a reaction rate.^[ref]

A proof outline comes from the reaction Gibbs free energy for dissolution:Here $R$ is the molar gas constant and $T>0$ the absolute temperature. If $Q_{\mathrm{sp}}>K_{\mathrm{sp}}$, the logarithm is positive, making the reverse transfer into solid favorable. At equality the bulk driving difference vanishes. Equality can also occur at the boundary where the last solid disappears, but it is not a universal constraint on every solid-free chemical solution.

A decimal significand lies in $[1,10)$, obtained by removing the integer power of ten from a positive number. Multiplication by a constant rotates the fractional logarithm modulo one. Invariance under every such rotation makes the fractional logarithm uniform and hence implies Benford law. This is an invariant significand distribution, not a proper scale-invariant distribution for the entire positive number.

At an unramified irregular singular point with distinct leading eigenvalues, a sectorial fundamental solution is a fundamental matrix whose entries are holomorphic functions asymptotic to a prescribed formal matrix on a sector of the logarithm cover. For competing cubic exponentials, the usual construction uses six overlapping sectors with basic angle $\pi /3$. A formal power series need not converge, and two sectorial realizations may differ by exponentially small terms across a sector boundary.

Take identical rigid spheres of radius $a$, particle density ${\rho }_p$ and displaced liquid density ${\rho }_f$, at uniform absolute temperature $T$. Let $z$ increase upwards from a reference layer and write $\Delta \rho ={\rho }_p-{\rho }_f>0$. The buoyancy correction gives the effective gravitational massThe actual grain mass is ${\rho }_{p}V$; the displaced liquid subtracts ${\rho }_{f}V$ from the effective gravitational mass. In a dilute, noninteracting particle suspension, the Boltzmann distribution therefore givesHere $n$ is number density and $k_B$ is the Boltzmann constant. Equivalently, the dilute osmotic pressure $\Pi =n{k}_{B}T$ supports the grains above it: $d\Pi /dz=-n{m}_{b}g$. Dividing by $n$ and integrating yields the same exponential function. No dynamic viscosity appears in this equilibrium argument. If ${\rho }_p<{\rho }_f$, the population instead increases upwards; at equal densities this gravitational model predicts uniform $n$.

A microscope counts a finite optical slab, not an infinitesimal plane. For equal viewed area $A$ and slab thickness $\delta$, an ideal uniform detection window centred at $z$ givesThus equal optical volumes preserve the exponential function count ratio exactly, even for finite $\delta$, provided the windows lie wholly inside the accessible chamber and have the same detection response. Unequal depth response or overlapping a boundary can corrupt this inference. Repeated observations estimate the mean count; closely spaced visits need not be independent because grains remain in view.

For an original chosen model, set $a=0.250\mu \mathrm{m}$, ${\rho }_p=1200\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $T=298\mathrm{K}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. These are illustrative inputs, not Jean Baptiste Perrin's measurements. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesIf equal observation effort gives expected count $400$ in the reference layer, the expected counts $20$, $40$ and $60\mu \mathrm{m}$ higher are approximately $214.3$, $114.8$ and $61.49$. Fractions describe expected values over observations, not fractional grains. Using the actual grain mass instead of its buoyancy-corrected value would make the inferred Avogadro constant six times too small for these chosen densities.

A semi-log plot uses a logarithmic scale on one axis and a linear scale on the other. Which axis is logarithmic matters. Plotting $x$ against $\log y$ makes positive exponential functions appear as straight lines; plotting $\log x$ against $y$ instead makes an affine function of a logarithm appear straight. The plotting operations change positions while retaining the original tick values.^[ref]

Take the original dimensionless mathematical model $y(t)=3e^{0.4t}$ for $0\le t\le 8$. It has exponential growth, with $y(0)=3$, $y(4)\approx 14.8591$, and $y(8)\approx 73.5976$. Taking a base-ten logarithm givesThe slope in these transformed positions is $0.4/\ln 10\approx 0.173718$, although the original relative growth rate $y^{\prime }/y$ is $0.4$. Equal steps of two in $t$ multiply $y$ by $e^{0.8}\approx 2.22554$. The corresponding ordinary gains increase, so the left plot bends while the right plot is straight.

One practical use is checking a proposed exponential distribution for failure times. For rate $\lambda >0$ and $t\ge 0$, its cumulative distribution function is $F(t)=1-e^{-\lambda t}$, so the reciprocal survival function $y=1/[1-F(t)]$ satisfies ${\log }_{10}y=\lambda t/\ln 10$. Estimated probabilities can therefore be placed on a logarithmic vertical axis and compared with a straight line. Agreement supports compatibility with the chosen statistical model; a roughly straight finite sample does not establish a unique physical mechanism.^[ref]

A first-order ordinary differential equation is separable if its explicit form factors as $y^{\prime }=a(t)b(y)$. On an interval where $b(y)\ne 0$, write $dy/b(y)=a(t)dt$ and take antiderivatives to obtain an implicit solution:An initial condition determines $C$ and selects a branch and domain. The useful simplification is that the two integrals each involve only one variable, even when the original equation is nonlinear.

For Newton's law of cooling, division by $y-T_a$ gives $dy/(y-T_a)=-kdt$. To keep the logarithm dimensionless, integrate relative to the initial excess temperature:When $y_0\ne T_a$, continuity preserves the sign of the excess temperature until any potential zero, giving $y-T_a=(y_0-T_a)e^{-k(t-t_0)}$. This formula never reaches zero at finite time. Taking its derivative verifies the ordinary differential equation, and substitution at $t_0$ verifies the initial condition.

Always check what division removed. Each zero $b(y_{\ast })=0$ gives a constant solution of a differential equation $y=y_{\ast }$. Cooling therefore also has the equilibrium point $y=T_a$, omitted by division but included in the final explicit formula. For the nonunique initial value problem $y^{\prime }=\sqrt{|y|}$, division by $\sqrt{|y|}$ only describes the moving pieces: a valid solution may remain zero for an arbitrary time before joining one. Separation is a method for finding solutions, not a uniqueness theorem.

For $T(n)=1$ at $n\le 3$ and $T(n)=2T(\lfloor n/2\rfloor {)}^2$ otherwise, put $d=\max (0,\lfloor {\log }_{2}n\rfloor -1)$ for positive integer $n$. Taking logarithms gives $S(n)=1+2S(\lfloor n/2\rfloor )$, hence $S(n)=2^d-1$ and $T(n)=2^{2^d-1}$. These constant-value plateaux have exponential envelopes $2^{\Theta (n)}$, but do not justify $\Theta (2^{n/2})$ for all integer inputs.

A steady state does not by itself give every fluctuating observable an entropy production symmetry. If the dimensionless total entropy production ${\Sigma }_t$ has the same-law finite relation and a large deviation principle with rate function $I$ for ${\Sigma }_t/t$, then, on its common domain,The first expression denotes leading exponential scaling; the second follows by taking the scaled logarithm of the probability ratio. A finite-state, autonomous Markov process with time-even states, positive bidirectional rates and a positive stationary distribution supplies a standard setting for a finite total-entropy production relation. The stochastic system entropy is then bounded, so its endpoint contribution does not change the long-time rate function of bath entropy production. ^[ref]

Unbounded state spaces can have endpoint fluctuations on the scale relevant to a large deviation principle. Dropping system entropy, changing the measured observable to heat, or changing the reverse preparation therefore requires a fresh argument. The exact finite integral fluctuation theorem, a same-law finite relation, and an asymptotic rate function symmetry are distinct claims.

For a long homogeneous circular wire of radius $a$, uniform volumetric heating $H$, constant thermal conductivity $\kappa$ and negligible end losses, the steady heat equation is $\kappa {\rho }^{-1}(\rho T^{\prime }{)}^{\prime }+H=0$. Regularity at the axis excludes a logarithm term, giving $T=T_s+H(a^2-{\rho }^2)/(4\kappa )$. If the surface obeys Newton's law of cooling, $-\kappa T^{\prime }(a)=\gamma (T_s-T_{\mathrm{air}})$, then $T_s=T_{\mathrm{air}}+Ha/(2\gamma )$. Here $\gamma$ has units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ and $H$ has units $\mathrm{W}{\mathrm{m}}^{-3}$. For uniform electric current $I$ and constant electrical conductivity $\sigma$, Joule heating gives $H=I^2/(\sigma {\pi }^2{a}^4)$.

Consider an unnamed model substance with molar latent heat $L_m=40.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ and saturation pressure $p_1=10.0\mathrm{kPa}$ at $T_1=300\mathrm{K}$. Stipulate the ideal-vapor Clausius-Clapeyron approximation throughout $280\le T\le 340\mathrm{K}$, constant $L_m$, and $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These parameters define an illustrative model, not a measured dataset or a fit to water. At $T_2=330\mathrm{K}$,At the reference state the gas molar volume is $R{T}_1/p_1=0.24942{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, and the local slope is $dp/dT\simeq 0.5346\mathrm{kPa}{\mathrm{K}}^{-1}$. The tangent estimate $p_2\approx 10.0+30(0.5346)=26.04\mathrm{kPa}$ is much lower than the integrated result: a local derivative should not be extrapolated over this whole interval. By contrast, over a small temperature increment, the fractional sensitivity is $d\ln (p/p_1)/dT=L_m/(R{T}_1^2)\simeq 0.05346{\mathrm{K}}^{-1}$.

Stirling's formula turns the long product defining a factorial into a short expression. For positive integers $n$, putThen $n!\sim S_n$ as $n\to \infty$. This is asymptotic equivalence: $S_n/n!\to 1$, so the relative error tends to zero. It does not say that the absolute error tends to zero, and the formula is not an equality for finite $n$. The Gamma function extends this approximation beyond integer inputs. ^[ref]

For $n=10$,The leading estimate is about $0.830\%$ low. Multiplying by the first correction factor giveswhich is about $0.003176\%$ low. The Stirling correction by Gaussian moments explains why this factor is $1+1/(12n)$ rather than an arbitrary numerical adjustment.

The usefulness is in ratios as well as in enormous numbers. Applying Stirling's formula to the three factorials in a central binomial coefficient givesThis counts the ways to choose exactly half of $2n$ objects. Equivalently, for $2n$ independent fair coin tosses, the binomial distribution assigns probability $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\sim 1/\sqrt{\pi n}$ to exactly $n$ heads. An exponentially large count becomes a simple probability scale after cancellation.

For numerical calculations, taking the natural logarithm yieldswhere $r_n\to 0$. Computing a logarithm of a large factorial avoids representing the factorial itself. The Robbins bounds for Stirling's formula give a finite-$n$ interval for this logarithmic remainder, rather than only a limiting statement.

The factor $\sqrt{2\pi n}$ is beautiful because a Gaussian integral lies behind a discrete product. The Gaussian peak behind Stirling's formula displays this connection through the Gamma integral and Laplace's method. A rigorous proof must also control the tails of the integral; the local expansion explains the geometry but is not by itself a complete proof.

James Stirling published Methodus differentialis in 1730, a treatise on summation and interpolation of infinite series. This setting connects the formula with a practical calculation problem: taking logarithms turns a long product into a sum, whose growth can be approximated. ^[ref]

Consider an original ideal model: $V_A=10.00\mathrm{mL}$ of hydrochloric acid at $c_A=0.100\mathrm{mol}{\mathrm{L}}^{-1}$, titrated with sodium hydroxide at $c_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$. Treat these numbers as exact model inputs. Assume complete dissociation of an ionic compound, complete strong acid ionization, additive volumes, uniform mixing, and activity coefficients equal to one. Use stipulated $K_w=10^{-14}$, approximately appropriate near $25{}^{\circ }\mathrm{C}$. These are calculated pH values, not measurements of these materials.

The chemical equation for neutralization isInitial acid inventory is $n_A=1.000\mathrm{mmol}$, so the equivalence point is $v=10.00\mathrm{mL}$. Added water does not change this inventory. But the final volume $V_A+v$, including the titrant, determines all concentrations.

Keep self-ionization of water even at the equivalence point. With $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, define the dimensionless acid excess $d$, hydronium level $h$, and hydroxide level $o$ byThe sodium cation and chloride anion inventories fix the electric charge balance $h-o=d$, while chemical equilibrium gives $ho=K_w$. Eliminating $o$ yields a quadratic equation $h^2-dh-K_w=0$ with exactly one positive root. For stable calculation on either side, writeThe second expression follows by multiplying by $s-d$; it avoids subtracting two nearly equal large quantities on the alkaline side. A logarithm acts on the dimensionless $h$, as required by the pH definition.

At $v=5.00\mathrm{mL}$, the excess is $0.500\mathrm{mmol}$ in $15.00\mathrm{mL}$, giving $d=1/30$ and pH $1.477121\dots$. At $v=15.00\mathrm{mL}$, excess hydroxide is $0.500\mathrm{mmol}$ in $25.00\mathrm{mL}$, giving $d=-1/50$ and pH $12.301030\dots$. Neglecting dilution would give the wrong values even far from the equivalence point.

Exactly at $v=10.00\mathrm{mL}$, $d=0$ and $h=o=\sqrt{K_w}=10^{-7}$, giving pH $7$ in this model. At $9.999\mathrm{mL}$ and $10.001\mathrm{mL}$, the same smooth expression gives pH $5.300835\dots$ and $8.699121\dots$. The familiar excess-only formulas would predict zero hydronium or zero hydroxide at the equivalence point; retaining self-ionization of water removes that failure. A steep curve is still continuous.

The sensitivity is especially revealing. Differentiating $h-K_w/h=d$ giveswhen $c_A=c_B$. At this model's equivalence point, the derivative is about $1.086\times 10^4$ pH units per millilitre. This is a local derivative, not a linear prediction over a whole drop. Real endpoint accuracy also depends on dosing, mixing, measurement uncertainty, temperature, extra reacting chemical species, and nonideal thermodynamic activities.

A centered random variable $X$ is sub-exponential with parameters $({\nu }^2,b)$ ifwhere $\nu ,b>0$. Unlike a sub-Gaussian random variable, the moment-generating-function bound is needed only near zero. Optimizing exponential Markov inequality over this restricted interval gives a two-sided tail bound of the formwith an absolute constant $c>0$. Small deviations therefore have a quadratic exponent and large deviations a linear exponent. Equivalently, up to constant changes in parameters, $\mathbb{E}{e}^{|X|/K}$ is bounded for some finite scale $K$.

For an exponential random variable $Y$ of mean one, $X=Y-1$ has moment-generating function $e^{-t}/(1-t)$ for $t<1$. Its logarithm is ${\sum }_{k\ge 2}{t}^k/k$, which is at most $t^2$ for $|t|\le 1/2$, proving a sub-exponential bound. The Centered square of a sub-Gaussian random variable is another example. The term here refers to exponential-tail concentration, and differs from the subexponential distribution defined by heavy-tail convolution asymptotics.