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In a population of $N$ playing rock paper scissors, let losers copy winners. With pair-selection events of rate $(N-1)/2$, the proportion process has jumps $(e_1-e_3)/N$, $(e_2-e_1)/N$, $(e_3-e_2)/N$ at rates $N{x}_1{x}_3$, $N{x}_2{x}_1$, $N{x}_3{x}_2$. Its fluid limit obeys ${\dot{x}}_i=x_i(x_{i-1}-x_{i+1})$, with indices interpreted cyclically. The product of numbers $x_1{x}_2{x}_3$ is conserved in this ordinary differential equation, whereas the finite chain eventually reaches a monochromatic absorbing state.

The Gamma function extends factorials to a continuous argument. Its defining Gamma integral iswhere $t^{z-1}=\exp ((z-1)\log t)$ uses the real natural logarithm for $t>0$. The Gamma function recurrence gives its meromorphic continuation to the complex plane, with simple poles at $0,-1,-2,\dots$ and no zeros. See Gamma function residue at a nonpositive integer and Gamma function has no zeros for proof outlines.^[ref]

For nonnegative integers $n$, the factorial identity has a shift of one:Thus $\Gamma (n)=(n-1)!$ for positive integers $n$. Half-integer arguments supply a different familiar constant:The first value follows from the Gaussian integral; the second follows from the Gamma function recurrence.^[ref]

The plot shows only positive real arguments; the gamma function also accepts complex numbers away from its poles. Matching the factorials alone does not select a unique interpolation, even if the Gamma function recurrence is imposed: $f(x)=\Gamma (x)\exp (\sin (2\pi x))$ has the same positive-integer values and the same recurrence relation. The Bohr-Mollerup theorem supplies the extra condition that selects the gamma function.

The useful idea is to replace a discrete product of numbers by an integral whose exponent varies continuously. The Gamma integral then evaluates whole families of integrals at once, provides the normalizing constant for the Gamma distribution in probability theory, and appears in physics, for example in the Bose integral. Its large-positive-argument growth reconnects this continuous description to factorials through Stirling's formula.^[ref]

Historically, Leonhard Euler studied the interpolation of factorials in his paper E19, written in 1729 and published in 1738. His integral over $(0,1)$ becomes the Gamma integral after a change of variables, with the same one-step shift in the argument.^[ref]

The geometric mean replaces a list of nonnegative real numbers by one repeated value that preserves their product of numbers. For $n\ge 1$ inputs,For two positive inputs, $G=\sqrt{x_1{x}_2}$ is their positive square root. The geometric mean of $1$ and $9$ is three: replacing $1\cdot 9$ by $3\cdot 3$ retains the same product. In contrast, their arithmetic mean five retains the sum $1+9=5+5$. The geometric mean is zero if any input is zero; the logarithm formula below requires every input to be positive. ^[ref].

Its usefulness becomes immediate for successive growth factors. Suppose a hypothetical positive quantity starts at $Q_0=1000$, is multiplied by $3/2$, then by $2/3$. It goes $1000\to 1500\to 1000$, so the geometric mean factor is $\sqrt{(3/2)(2/3)}=1$: a constant factor of one for both steps gives the same endpoint. The arithmetic mean factor is $13/12$; repeating it would give $1000(13/12{)}^2\approx 1173.61$, a different endpoint. A mean of factors summarizes multiplicative change; subtracting one afterwards gives the equivalent change per step. Equal-duration steps are necessary to interpret this as change per common time interval.

For positive inputs, logarithms turn multiplication into addition, soThus the geometric mean is an arithmetic mean in logarithmic coordinates, followed by the exponential function. For positive measurements with a common unit, one can use dimensionless ratios to a fixed reference $x_0$: $G=x_0\exp (n^{-1}{\sum }_i\log (x_i/x_0))$. Changing the common unit rescales $G$ by the same factor as the inputs. As a check, doubling one input and halving another retains $G$ while usually changing the arithmetic mean.

The arithmetic-geometric mean inequality gives $G\le A$. Apply that same inequality to the multiplicative inverses $1/x_i$ to obtain $H\le G$, where $H$ is the harmonic mean. Consequentlyfor positive inputs, with equality throughout exactly when all inputs agree. For $30$ and $60$, the three values are $H=40$, $G=30\sqrt{2}\approx 42.4264$, and $A=45$. They preserve, respectively, the sum of inverses, the product, and the sum. No one of these summaries preserves all three for unequal inputs.

For positive endpoints $a,b$, halfway along a logarithmic scale meansHere $\beta >1$ is the base. Equivalently, using the natural logarithm, $2\ln (m/x_0)=\ln (a/x_0)+\ln (b/x_0)$. The product of numbers identity for logarithms gives $m^2=ab$, so the midpoint is the geometric mean $\sqrt{ab}$. The two endpoints lie at equal multiplicative distances because $m/a=b/m$.

For $a=10$ and $b=100$, this midpoint is $\sqrt{1000}\approx 31.6228$, giving a factor of $\sqrt{10}$ on each side. The arithmetic mean $55$ is halfway on a linear scale, but its two ratios are $55/10=5.5$ and $100/55\approx 1.81818$. Thus reading halfway between the marks $10$ and $100$ as $55$ on a logarithmic scale gives the wrong value.

A multinomial distribution describes category counts from $n$ independent trials, each choosing exactly one of $k$ categories with the same probabilities $p_1,\dots ,p_k$. Require $n\ge 0$ to be an integer, $p_i\ge 0$, and ${\sum }_i{p}_i=1$. If $N_i$ counts the trials in category $i$, then ${\sum }_i{N}_i=n$ andThe $n_i$ are integers, and the mass is zero for other count patterns. A category with $p_i=0$ has count zero almost surely, with $0^0=1$ in this formula. For a specified ordered trial sequence, independence gives the product of numbers ${\prod }_i{p}_i^{n_i}$; the multinomial coefficient counts the different sequences with that same count pattern.

For four trials with category probabilities $(1/2,3/10,1/5)$, the count pattern $(2,1,1)$ can occur in $4!/(2!1!1!)=12$ orders, each of probability $(1/2{)}^2(3/10)(1/5)=0.015$. Thus its total probability is $12(0.015)=0.18$. This is a constructed independent-trial example, distinct from drawing four objects without replacement from a ten-object population of composition $(5,3,2)$, where the multivariate hypergeometric distribution gives $2/7$.

The diagram shows all fifteen possible count patterns. Coordinates $(N_1,N_2)$ determine $N_3=4-N_1-N_2$; the triangle is a constraint on integer counts, not a continuous probability density. The mode is $(N_1,N_2,N_3)=(2,1,1)$, where the probability mass function is largest, but the expected values $(2,1.2,0.8)$ are not a possible count pattern.

Each single count has a binomial distribution, because recording category $i$ versus all other categories reduces every trial to a Bernoulli trial. Consequently,These statements about the joint count model are summarized in Aditya Guntuboyina's lecture notes.^[ref]

Independent trials do not make the resulting counts independent random variables. Within one trial the indicator random variables for two distinct categories cannot both equal one, so their product is zero and their covariance is $-p_i{p}_j$. Summing over independent trials gives $-n{p}_i{p}_j$. In the four-trial example, the variances are $(1,0.84,0.64)$ and the pairwise covariances are $-0.6,-0.4,-0.24$. Knowing that all four trials were in the first category fixes the other two counts at zero.

The attraction is that one formula describes any fixed-total list of category counts while preserving the constraint that they add up to $n$. It underlies multinomial likelihood calculations and uniform balls-in-bins allocation. The multinomial aggregation rule lets us merge categories without leaving this family of probability distributions. Allowing independent category counts with an unfixed total instead leads to a different model, related through Poisson-multinomial conditioning.

Multiplication combines two factors into their product of numbers. For a natural number $n$, $nx$ is the sum of $n$ copies of $x$; with no copies the result is zero. This repeated-addition interpretation directly explains integer grouping, while multiplication of arbitrary real numbers also describes scaling by noninteger factors.

A rectangle of side lengths $3$ and $6$ has area $18$. Split the second side into lengths $2$ and $4$. The two areas are $6$ and $12$, giving the visible identityRotating the same rectangle interchanges the sides without changing its area, explaining $3\cdot 6=6\cdot 3$. These pictures illustrate distributivity and commutativity for ordinary positive numbers; matrix multiplication and other operations require their own rules.

For a fractional example, take $3/2$ of a quantity and then $2/3$ of that result: the combined scale is $(2/3)(3/2)=1$. The two fractions are multiplicative inverses. In the unit square on the right, the overlap of a strip occupying $3/4$ of the width with one occupying $2/3$ of the height contains six of twelve equal cells. Its area is $(3/4)(2/3)=1/2$. The product stays exact even though the decimal expansion of $2/3$ repeats. ^[ref]

Fix a prime number $p$. Write a nonzero rational number as $x=p^{k}a/b$, where $k\in \mathbb{Z}$ and the nonzero integers $a,b$ are not divisible by $p$. The exponent $k$ is uniquely determined. Its P-adic valuation is this signed exponent $v_p(x)=k$: equivalently, for any nonzero integer numerator $A$ and denominator $B$,Factors of $p$ in the numerator decrease the p-adic absolute value; factors in the denominator increase it. For $p=2$ the computations are:

This is an absolute value on a field measuring divisibility, rather than the magnitude on a real number number line. It satisfiesThe proof comes directly from P-adic valuations: exponents add in a product of numbers, and extracting a common power of $p$ from a sum gives $v_p(x+y)\ge \min \{v_p(x),v_p(y)\}$. If the two P-adic valuations differ, the remaining term with smaller exponent cannot cancel modulo $p$, so equality holds. Thus the p-adic absolute value is a Non-Archimedean absolute value. ^[ref]

The metric $d_p(x,y)=|x-y{|}_p$ makes two rational numbers close when their difference contains a high power of $p$. This packages successive modular congruences into a numerical error estimate, allowing convergence in a metric space and convergent series to encode exact arithmetic information.

Use coordinates $1,2,3$ for paper, scissors and stone. Let $X^N$ take values in the lattice simplex $S_N=\{(n_1,n_2,n_3)/N:n_i\ge 0,{\sum }_i{n}_i=N\}$. A uniformly selected unordered pair consists of types $i$ and $j$, $i\ne j$, with probability $2n_i{n}_j/(N(N-1))$. Losing copies the winner, so the possible jumps of the continuous-time Markov chain areTies make no change. For arbitrary event rate $\lambda$, the respective transition rates are $2\lambda N{x}_1{x}_3/(N-1)$, $2\lambda N{x}_2{x}_1/(N-1)$ and $2\lambda N{x}_3{x}_2/(N-1)$. ChooseThen the rates are exactly $N{x}_1{x}_3$, $N{x}_2{x}_1$, $N{x}_3{x}_2$. The infinitesimal generator of this cyclic imitation chain iswhere $\mathcal{J}=\{e_1-e_3,e_2-e_1,e_3-e_2\}$ and the three $\beta$ functions are, in that order, $x_1{x}_3,x_2{x}_1,x_3{x}_2$. Its drift is precisely

Here is a precise fluid limit with its error control. Suppose $X_0^N\to x_0$ in probability, and let $x$ solve $\dot{x}=F(x)$, $x(0)=x_0$. For every finite $T$,Indeed, the martingale decomposition of a density-dependent jump process givesThis follows by compensating each transition count by its integrated rate; the compensated counts are square-integrable martingales, with variance equal to the expected compensator. Since $\Vert \ell {\Vert }^2=2$ and $x_1{x}_2+x_2{x}_3+x_3{x}_1\le 1/3$ on the simplex, the Doob L2 maximal inequality, applied coordinatewise, givesThe polynomial drift is Lipschitz on the compact simplex, with some finite constant $L$. It preserves the sum of the coordinates, and each coordinate solves an equation of the form ${\dot{x}}_i=x_i{h}_i(x)$, so nonnegativity is preserved. Thus the ordinary differential equation stays in the simplex and exists for all time. The Gronwall inequality giveswhich proves the claim by Markov's inequality. This proves the required version directly, rather than invoking an unspecified approximation theorem.

For an interior trajectory, differentiating the logarithm of the product of numbers givesThe same identity for the product itself holds on the boundary, so $x_1(t)x_2(t)x_3(t)$ is conserved. A strictly positive initial product keeps the deterministic trajectory away from the boundary. Except at $(1/3,1/3,1/3)$ these interior trajectories are periodic: the strictly concave function ${\sum }_i\log x_i$ has smooth compact level curves surrounding its unique maximum, and the vector field is nonzero and tangent to each such curve.

The finite Markov chain behaves differently at long times. For $P(x)=x_1{x}_2{x}_3$, the change in $P$ during the first jump isThe other two terms are cyclic permutations. Their first-order contributions cancel, leaving the exact product decay in a cyclic imitation chain identityFor $N\ge 3$, while all three counts are positive, their product is at least $N-2$, so $P(X_t^N)\ge (N-2)/N^3$. A missing type cannot reappear. If $\tau$ is the first loss of a type, thenOnce only two types remain, one always wins, so its count increases until an absorbing monochromatic vertex is reached, almost surely in finite time. The cases of one or two initially present types follow the same argument without the first stage. Every finite chain eventually becomes monochromatic, whereas an interior deterministic solution retains its positive product. The fluid limit holds on each fixed finite interval; it does not justify exchanging the limits $N\to \infty$ and $t\to \infty$.

For the cyclic imitation chain, its infinitesimal generator satisfies ${\mathcal{L}}_N(x_1{x}_2{x}_3)=-3x_1{x}_2{x}_3/N$. Therefore $\mathbb{E}[X_1^N(t)X_2^N(t)X_3^N(t)]=e^{-3t/N}\mathbb{E}[X_1^N(0)X_2^N(0)X_3^N(0)]$. While all three types remain, the product of numbers is at least $(N-2)/N^3$ for $N\ge 3$, so this identity proves that one type is lost almost surely. The remaining pair has a fixed winner, which eventually occupies the whole population. The conserved product in the deterministic fluid limit therefore describes fixed finite horizons, not the chain's infinite-time behavior.

Take independent random variables $U_n$, each with the continuous uniform distribution on $[0,1]$, and set $A_n=\{U_n\le p_n\}$. Then $\mathbb{P}(A_n)=p_n$, and the events are mutually independent events. The product probability space supplies a model for the entire infinite sequence; a finite list of simulated values is only an illustration.

For $p_n=1/(n+1{)}^2$, the p-series is summable:The Basel problem supplies the total sum, and the integral test for convergence gives the displayed tail bound. The First Borel-Cantelli lemma makes the limsup of events have probability zero. In fact, the tail bound implies that the probability of even one alarm at or after $m$ is at most $1/m$; for $m=100$, it is at most $0.01$. This bound remains valid for dependent events with the same individual probabilities.

For $p_n=1/(n+1)$, the harmonic series diverges, so the Second Borel-Cantelli lemma makes the limsup of events have probability one. There is also an exact finite-horizon calculation using independent events:For any fixed $m$, this finite product of numbers tends to zero as $M\to \infty$. Every infinite tail therefore contains an alarm almost surely, even though the individual probabilities tend to zero.