Only one curve through the factorials bends the right way

A rectangle cut by a curve integrated by parts over and over and found the factorials among the areas: the area under $x^n e^{-x}$ over the whole half-line is $n!$. Because the area makes sense when $n$ is not a whole number, it extends the factorial to every positive real number — Euler’s gamma function, $\Gamma(s) = \int_0^\infty x^{s-1} e^{-x}\,dx$, with $\Gamma(n+1) = n!$. The extension gives $(\tfrac12)! = \sqrt\pi/2$, puts the factorial into the volumes of balls in every dimension, and is so useful that it is easy to forget to ask whether it is the only possible one.

It is not. Infinitely many smooth curves pass through every factorial. So the question is what makes Euler’s choice the factorial of a fraction rather than a factorial. The answer, found by Harald Bohr and Johannes Mollerup in 1922, is a single condition on the shape of the curve: its logarithm must bend upwards everywhere. This essay draws the impostors that condition rules out, measures how each one fails it, and runs the squeeze that proves nothing else survives.

Values do not decide a curve #

The first impostor is the easiest to build. Any finite list of points is passed through by a polynomial, and the polynomial of degree six through $0!, 1!, \dots, 6!$ matches the factorial at those seven points exactly. Between them it does what it likes: at $x = \tfrac12$ it gives $-3.58$, a negative factorial, while Euler’s integral gives $0.8862$. A polynomial through the first hundred factorials would do something different again, and nothing about the values alone prefers any of them.

The second impostor is more insidious. Multiply $\Gamma(x+1)$ by $1 + \tfrac12 \sin^2 \pi x$. The extra factor is exactly 1 at every whole number, since $\sin \pi n = 0$, so the product passes through every factorial. And because $\sin^2 \pi x$ repeats with period 1, the product also obeys the factorial’s own rule, $f(x+1) = (x+1)\,f(x)$, at every real $x$ — not just at whole numbers. At $x = \tfrac12$ it gives $1.3293$. It is as good a factorial as Euler’s by every test the factorial itself can supply.

Values and the rule together therefore leave infinitely many candidates. The figure that divides each candidate by $\Gamma(x+1)$ shows exactly how much freedom is left.

Any function that obeys $f(x+1) = (x+1) f(x)$ and is never nought is $\Gamma(x+1)$ times a function of period one: divide the rule for $f$ by the rule for $\Gamma$ and the ratio $p(x)$ satisfies $p(x+1) = p(x)$. The orange ratio repeats exactly, $1 + \tfrac12\sin^2\pi x$, over and over. The polynomial only matches values, and its ratio drifts: $-4.04$, $2.07$, $0.71$, $1.10$, $0.95$, $1.05$ at the half-integers from $\tfrac12$ to $5\tfrac12$. The rule turns the question “which curve?” into the question “which periodic factor?”, and the answer has to come from somewhere else.

This is the same situation as a function that adds and is nowhere a line, where the rule $f(x + y) = f(x) + f(y)$ has the obvious solutions $f(x) = cx$ and also, with the axiom of choice, wild solutions that are nowhere continuous. A rule about how a function behaves under a shift or a sum pins it down only on a skeleton — the whole numbers, or the rationals — and between the bones anything consistent with the rule is allowed. What rescues the obvious solution in both cases is a condition of regularity. For the additive rule, continuity at a single point is enough. For the factorial’s rule, continuity is not enough — the wiggled curve is perfectly smooth — and something stronger about shape is needed.

The logarithm that bends upwards #

The condition Bohr and Mollerup found concerns the logarithm of the curve.

The logarithm of $\Gamma$ bends upwards everywhere. Its second derivative is the sum $\sum_{k \ge 0} 1/(x+k)^2$, which is positive at every point, and so every chord of the curve lies above it: $\log \Gamma$ is a convex function, and $\Gamma$ is called log-convex. The wiggled curve’s logarithm is $\log\Gamma(x+1)$ plus $\log(1 + \tfrac12 \sin^2 \pi x)$, and the second term bends down hard just beyond each whole number; over 54% of the range drawn, the sum bends downwards. It is not log-convex.

Log-convexity is a natural condition, not an arbitrary one. It says that the ratio $f(x+h)/f(x)$, for a fixed step $h$, never decreases as $x$ grows: the factorial multiplies by $n + 1$ to get from $n!$ to $(n+1)!$, and those multipliers grow, so a curve whose growth rate ever falls back has done something the factorials never do. It is also the property that makes the average of a curve and the curve of an average compare the right way for the logarithm, which is where many of the gamma function’s inequalities come from. Bohr and Mollerup’s theorem is that it is enough:

The gamma function is the only function on the positive reals that is positive, log-convex, takes the value 1 at 1, and satisfies $f(x+1) = x f(x)$.

Emil Artin made the theorem the definition, in a short book of 1931 that develops every property of $\Gamma$ from those three conditions rather than from the integral. The integral then becomes a formula for the unique function the conditions describe, rather than its definition.

Why the logarithm, and not the curve #

It is tempting to ask for something simpler: that the curve itself bend upwards, with no logarithm. $\Gamma(x+1)$ does bend upwards for every positive $x$, and a wiggled curve of size one half visibly does not. But a small wiggle passes this weaker test everywhere. For $a = 0.03$ the curve $\Gamma(x+1)(1 + a\sin^2\pi x)$ has a positive second derivative at every point checked from $x = 0$ to $x = 60$, the closest call being near $x = \tfrac12$, and beyond that its growth makes failure impossible: the curvature of the factorial, relative to its size, grows like the square of $\log x$, while the wiggle’s relative curvature stays below $2\pi^2 a$. For $a = 0.01$ and $a = 0.001$ the margin is wider still. So there are infinitely many convex curves through the factorials that obey the factorial’s rule, and plain convexity cannot choose among them.

The logarithm makes the difference because it removes the growth. A curve that grows like $n!$ bends upwards so strongly that a small periodic factor is invisible to the curve’s own curvature; the logarithm of the curve grows only like $n \log n$, and the curvature of that straightens out as $x$ grows, so the periodic factor is eventually exposed. Asking for convexity of the logarithm is asking for convexity at the right scale — the scale at which the factorial’s growth has been divided out and only its rate of growth is being examined. That rate is what the multipliers $1, 2, 3, \dots$ describe, and the condition says the rate never falls back.

The straightening of $\log n!$ is the same fact that Stirling’s approximation expresses in numbers: $\log n!$ is $n \log n - n$ plus terms that grow ever more slowly, so over any stretch of fixed length it is almost a straight line. And the chords and slopes that the squeeze compares are the supporting lines of a convex function, the objects a line under every point builds every convex curve from: a log-convex function is one whose logarithm lies above each of its tangent lines, and the squeeze is just two of those tangent lines and a chord, read off at the factorials.

Every wiggle fails somewhere #

The wiggle in the figures was large, and failed log-convexity everywhere. A small enough wiggle seems as though it ought to survive: the logarithm of $\Gamma$ bends upwards with a definite curvature, and a tiny periodic disturbance should not be able to overturn it.

For small $x$ it does survive. A wiggle of size $a = 0.0001$ keeps the logarithm convex all the way to $x \approx 512$. But it fails there, and every wiggle fails somewhere. The reason is a race between two curvatures. The curvature of $\log\Gamma(x+1)$ is about $1/x$ for large $x$, so the logarithm straightens out as $x$ grows — over any stretch of fixed length, $\log n!$ becomes nearly a straight line. The wiggle’s curvature does not shrink: $\log(1 + a\sin^2\pi x)$ bends down by up to about $2\pi^2 a$ just beyond each whole number, at every $x$. The two cross where $1/x \approx 2\pi^2 a$, which is the estimate drawn as the line, and halving the wiggle doubles how far out it survives — the measured slope is $-0.99$.

So log-convexity on a bounded interval would not single out $\Gamma$; small wiggles pass it. Log-convexity on the whole half-line does, because the condition grows stricter as the logarithm straightens, and eventually no periodic factor other than the constant 1 is allowed. Any proof of the theorem must use the condition far out along the half-line, and the one that does is a squeeze.

The squeeze that proves it #

Suppose $f$ is positive, log-convex, has $f(1) = 1$ and satisfies $f(x+1) = x f(x)$. Then $f(n) = (n-1)!$ at every whole number. Fix $0 < x \le 1$ and a whole number $n \ge 2$, and compare three slopes of the convex function $\log f$: over $[n-1, n]$, over $[n, n+x]$, and over $[n, n+1]$. Convexity puts them in increasing order, and the first and last are $\log(n-1)$ and $\log n$. That gives

$$(n-1)^x\,(n-1)! \;\le\; f(n+x) \;\le\; n^x\,(n-1)!,$$

and the rule carries $f(n+x)$ back to $f(x)$ by dividing by $x(x+1)\cdots(x+n-1)$.

The two bounds differ by the factor $(n/(n-1))^x$, which tends to 1 as $n$ grows. So for every $n$ the function $f$ is trapped in a band around a specific value, the band shrinks to nothing, and $f(x)$ is determined: it is the common limit,

$$f(x) = \lim_{n\to\infty} \frac{n^x\,(n-1)!}{x(x+1)\cdots(x+n-1)},$$

which is Gauss’s product formula for the gamma function. It is also, in another form, how the function first appeared: Euler’s letters to Christian Goldbach in 1729 introduced the extended factorial as an infinite product of exactly this kind, and the integral came afterwards. The squeeze explains why the product and the integral had to agree — both are log-convex, both obey the rule, and there is room for only one such function. Since $\Gamma$ satisfies all three conditions, $f = \Gamma$ on $(0, 1]$, and the rule extends the agreement to every positive $x$. The whole proof is one comparison of slopes and one limit, and it uses the condition at $n$, as far out as needed — exactly where the wiggles fail.

A squeeze is also a computation #

The bounds are explicit, so the proof is also an algorithm. At $x = \tfrac12$ the function’s value is $\Gamma(\tfrac12) = \sqrt\pi$, and the squeeze traps it.

With $n = 2$ the interval is 1.333 to 1.886; with $n = 16$, 1.7296 to 1.7864; with $n = 1{,}024$, 1.771804 to 1.772670, around $\sqrt\pi = 1.772454$. The width shrinks like $\sqrt\pi/(2n)$, so the squeeze is a slow way to compute $\sqrt\pi$ — each extra digit costs ten times as many factors — but it is a proof and a computation at once, with nothing used beyond the three conditions. A squeeze of the same shape — each area trapped between its two neighbours — is what produced Wallis’s product for $\pi$ from the powers of a sine in the essay that found the factorials among areas, and the agreement is not a coincidence: Wallis’s product is Gauss’s product formula evaluated at $x = \tfrac12$.

A faster route to $\sqrt\pi$ exists — it is the area under the bell curve, squared and computed in polar coordinates — and the fact that the slow squeeze gives the same number is a check that the three conditions really do describe Euler’s integral rather than some other function that happens to match the factorials.

Other ways to pick the factorial #

Log-convexity is not the only condition that singles out $\Gamma$. Helmut Wielandt proved in 1939 that among functions of a complex variable, $\Gamma$ is the only one that is analytic for positive real part, satisfies the rule and $f(1) = 1$, and stays bounded on the strip $1 \le \operatorname{Re} z \le 2$. The wiggled curve fails that condition too: $\sin^2 \pi z$ grows exponentially away from the real axis, so its periodic factor is unbounded on the strip. Again a regularity condition — this time growth in the complex plane rather than bending on the real line — rules out every periodic factor but the constant.

There are also extensions that deliberately give up the rule. Jacques Hadamard wrote down in 1894 a function that agrees with $(n-1)!$ at every positive whole number and is analytic in the entire complex plane, with no poles at the negative integers — something $\Gamma$ cannot do, since $\Gamma$ has a pole at every non-positive integer. Hadamard’s function does not satisfy $f(x+1) = x f(x)$ away from the whole numbers. The choice between it and $\Gamma$ is a choice between keeping the rule and keeping analyticity everywhere, and mathematics has overwhelmingly kept the rule, because the rule is what the factorial is for: the integrals of areas undoing slopes, the counting formulas and the volumes all use the step from $n$ to $n+1$.

The gamma function also has a property that marks it as transcendental in a strong sense. Otto Hölder proved in 1887 that it satisfies no algebraic differential equation — no polynomial relation among $\Gamma$ and finitely many of its derivatives. Log-convexity picks it out among extensions of the factorial; Hölder’s theorem says that, once picked, it is not reachable from the familiar functions by any equation of that kind.

What the figures cannot show #

The figures show finitely many impostors and finitely many values of $n$. The ratio figure shows that two particular candidates differ from $\Gamma$ by a periodic factor or by something worse; it does not show that every candidate obeying the rule is $\Gamma$ times a periodic factor, which is a two-line argument rather than a picture. The failure figure measures where eight particular wiggles stop being log-convex; the claim that every non-constant periodic factor fails somewhere is the content of the theorem and is proved by the squeeze, not by the measurements. And the squeeze figure shows the bounds for three values of $n$ on part of an interval; the limit, which is the proof, is not something a figure can contain.

There is also something the figures quietly assume. Every curve here is computed from the gamma function itself — the impostors are built by multiplying it by something — so the pictures cannot show what a log-convex extension of the factorial would look like if one were found by some other route. The theorem says there is nothing to find, and the pictures are consistent with that, which is all a picture can be.

Still open: conditions that work for other sequences #

For the factorial, the question is closed: three conditions, one function, a two-line proof by squeezing. The same question can be asked of any sequence defined by a rule, and for many of them it is not settled which regularity condition picks out the natural interpolation, or whether one exists. For products of the form $f(x+1) = g(x) f(x)$ with $g$ log-concave, theorems of the Bohr–Mollerup type have been proved by Wolfgang Krull and others, and they single out a unique log-convex solution; for faster-growing sequences such as the superfactorials $1!\,2!\cdots n!$, the corresponding extension — the Barnes G-function — is singled out by a stronger condition, convexity of a higher derivative of its logarithm, and a general account of which condition fits which rule is still being worked out.

A sharper question is quantitative. Log-convexity is a strong condition, and Wielandt’s boundedness is a different strong condition; how weak a condition can be and still force the gamma function — somewhere between continuity, which is too weak, and log-convexity — has been asked in several forms, with partial answers that depend on what kind of condition is allowed.