The curve that is its own slope

Most named constants are answers to a question anyone would ask. $\pi$ can be measured off a circle with string; $\sqrt 2$ is the diagonal of a unit square, which is how the Greeks ran into it. $\pi$ is how many diameters fit around a circle. $\sqrt2$ is the diagonal of a unit square. Both come from measuring something.

$e = 2.71828\ldots$ is not like that. Nobody measured anything and got 2.71828. It is the answer to a question that has to be asked before it can be answered — like the question of which cutting angle gives a parabola, it only exists once someone has set the problem up properly, and the question is this: is there a curve that grows at exactly the rate of its own height?

Why any base almost works #

Every exponential curve $y = b^x$ has a slope proportional to its height. That much is easy and follows from the algebra: the slope at $x$ works out to $k \cdot b^x$, where $k$ is a constant that depends only on $b$ and not on $x$.

The proportionality is automatic. What is not automatic is the constant.

For $b = 2$, the constant is about $0.693$: the curve $2^x$ climbs at roughly seven-tenths of its own height. For $b = 3$ it is about $1.099$ — steeper than its height. Somewhere between 2 and 3, the constant passes through exactly 1.

The two pictures bracket the answer, which is a better reason to believe it exists than any amount of algebra. The constant $k$ moves continuously as the base does — it is $\ln b$, though that is the conclusion rather than the starting point — and it is below 1 at $b = 2$ and above 1 at $b = 3$. Something continuous that is too small here and too large there is exactly 1 somewhere between.

That crossing point is $e$. It is defined by the property, not the other way round: $e$ is the number whose exponential curve has proportionality constant exactly 1.

Which sounds like a self-serving definition until the consequences pile up. Because $k = 1$ for base $e$ and $k \ne 1$ for every other base, calculus done in base $e$ has no stray constants in it and calculus done in any other base is full of them. The derivative of $e^x$ is $e^x$. The derivative of $2^x$ is $2^x \ln 2$. Every appearance of another base drags a conversion factor along, and that factor is a logarithm to base $e$ — which is why those logarithms get called natural.

The naturalness is not aesthetic. It is that one particular base makes the bookkeeping vanish, and the others do not.

Compound interest arrives at the same number #

The other standard route to $e$ looks entirely unrelated. Invest £1 at 100% annual interest. Paid once at the end of the year, that gives £2.

Pay it twice a year instead, 50% each time. The second payment earns interest on the first half’s interest, so the total is $(1.5)^2 = 2.25$. Four times a year: $(1.25)^4 = 2.4414$. Monthly: $2.6130$. Daily: $2.7146$.

The totals increase every time the compounding gets finer, which is intuitive: interest that arrives sooner starts earning sooner. What is not intuitive is that they do not increase without limit. Compounding hourly, per second, per nanosecond — the total creeps toward $2.71828\ldots$ and stops.

$$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$$

Two quantities are fighting here. The base $(1 + 1/n)$ shrinks toward 1, and raising 1 to any power gives 1. The exponent $n$ grows without bound, and raising anything above 1 to an unbounded power gives infinity. The limit is neither 1 nor infinity but a specific finite number in between, which is the sort of outcome that ought to be suspicious and turns out to be exactly right.

The connection to the tangent-line story is direct. Continuous compounding means growth proportional to current size at every instant — money earning interest with no waiting. Which is the same condition as the curve whose slope equals its height. The banker and the geometer are describing one situation.

The series that computes it #

There is a third description, and it is the one used to actually calculate the digits:

$$e = 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \cdots = \sum_{k=0}^{\infty}\frac{1}{k!}$$

The factorials in the denominators make this converge extremely fast — ten terms already give seven correct decimal places, which is far better behaved than the compound interest limit, where a million compounding periods still only gets about five.

This series is where the self-slope property becomes almost visible. Differentiate it term by term: $\frac{x^k}{k!}$ becomes $\frac{x^{k-1}}{(k-1)!}$, which is the term before it. So the whole series maps to itself, shifted, which is to say it maps to itself. The factorials are exactly the weights that make the shift into an identity.

Euler introduced the symbol $e$ in the 1720s, and it is worth saying that he almost certainly did not name it after himself. The plausible explanations are that $a$, $b$, $c$ and $d$ were already busy, or that $e$ stood for exponential.

What the digits actually cost #

Three descriptions of the same number, and they are not interchangeable once someone has to produce digits from them.

The series is extravagantly good. Its error after $m$ terms is smaller than the first term omitted, so the cost of one more decimal place falls as the calculation proceeds: five terms give two places, ten give seven, twenty give eighteen. The factorial in the denominator is doing something no polynomial rate can match — each term is smaller than the last by a factor that itself keeps growing.

The compound-interest limit is atrocious. Its error shrinks like $1/n$, which means a correct digit costs ten times the work of the digit before it. A million compounding periods buys about six places. Set against the series’ twenty terms for eighteen places, the two are not in the same trade.

That much is a statement about mathematics. What happens next is a statement about machines, and it is worse than slow.

Evaluated in ordinary double-precision arithmetic, $(1 + 1/n)^n$ does not merely converge slowly toward $2.71828\ldots$ — it converges for a while, turns around, and walks away. Near $n = 10^8$ the computed value is right to about seven places. By $n = 10^{16}$ the quantity $1 + 1/n$ has no representation distinct from $1$: the ratio is smaller than the gap between $1$ and the next representable number, so the addition returns $1$ exactly, and the answer collapses to $1^n = 1$. The formula that defines $e$ evaluates, at large enough $n$, to a number that is not close to it.

Nothing is wrong with the limit. What is wrong is the assumption that a mathematically convergent expression is a computational recipe. Two operations here fight each other — an addition that loses the small quantity, and an exponentiation that would have amplified it — and the arithmetic decides the fight before the limit gets a chance to.

This is the same lesson that rectangles under a curve teach in a different costume, and the same one that governs estimating $\pi$ by dropping needles, where the error falls only as $1/\sqrt{n}$ and four thousand throws buy two decimal places. A definition says what a number is. It carries no promise about what it costs to see.

The same factorials prove it is irrational #

The series is extravagant in a second way, and it belongs beside the first, because it turns the cost calculation into a theorem.

Suppose $e$ were rational — say $e = p/q$ for whole numbers $p$ and $q$. Multiply the series through by $q!$:

$$q!\,e = \underbrace{q!\sum_{k=0}^{q}\frac{1}{k!}}_{\text{an integer}} \;+\; \underbrace{q!\sum_{k=q+1}^{\infty}\frac{1}{k!}}_{\text{the tail}}.$$

The first sum is a whole number, because $q!/k!$ is a whole number for every $k \le q$. The left-hand side is a whole number too, since $q!\,p/q = (q-1)!\,p$. So the tail has to be one as well.

It is not, and this is where the factorials earn their keep. The tail is

$$\frac{1}{q+1} + \frac{1}{(q+1)(q+2)} + \cdots \;<\; \frac{1}{q+1} + \frac{1}{(q+1)^2} + \cdots = \frac{1}{q},$$

and it is plainly greater than zero. So it lies strictly between $0$ and $1/q$, and no whole number is there. Checked directly: $q!\,e$ minus the integer below it comes out at $0.2388$ when $q = 4$, against a bound of $0.25$; at $q = 7$ it is $0.1404$ against $0.1429$. Always inside, and squeezing tighter as $q$ grows.

That is Fourier’s argument from 1815, and the whole of it fits in a paragraph precisely because the terms fall factorially. It is worth setting beside $\pi$, which has no comparable series, and whose irrationality Lambert had to reach in 1761 by way of the continued fraction of the tangent — a substantially harder argument about a number that is in every other respect the more familiar of the two. The property that makes $e$ cheap to compute and the property that makes it easy to prove irrational are the same property, which is how fast $1/k!$ shrinks.

The condition the picture is quietly using #

The tangent-line property is not the only characterisation on offer, and the most natural-looking one has a hole in it.

An exponential is often introduced as the function that turns addition into multiplication: $f(x + y) = f(x)\,f(y)$. Growth over two years is growth over one year, twice — the defining behaviour of anything that compounds. It is tempting to conclude that this single equation forces $f(x) = b^x$ for some base.

It does not. The equation pins the function down on the rationals and then stops. Choosing $f(1) = b$ forces $f(2) = b^2$, $f(1/2) = \sqrt b$, and generally $f(p/q) = b^{p/q}$ — every rational exponent is determined. The irrationals are free, and there are solutions that assign them wildly, producing functions satisfying the equation everywhere while being unbounded on every interval, graphs so badly behaved they are dense in the whole plane. Such a function has no tangent lines anywhere, because it is not continuous anywhere.

These objects cannot be drawn, and that is not a limitation of the draughtsman. Constructing one requires the axiom of choice, and no explicit example exists; they are known to be there without ever being exhibited. The mildest assumption — continuity at a single point, or measurability, or boundedness on some interval however short — kills all of them at once and leaves exactly the exponentials.

So the picture is doing more work than it appears to. Drawing a smooth curve through the constraint is not an illustration of the functional equation; it is the functional equation plus a regularity assumption strong enough to eliminate a whole family of monsters. The assumption is nearly always true of anything a physical process produces, which is why it is nearly always left unstated, and why it is easy to forget it is an assumption at all.

What the picture cannot show #

The tangent-line figure is a claim about every point, and it draws four. Nothing in it rules out a curve that satisfies slope-equals-height at four places and fails between them — and that gap is not decorative, because the property at every point is what makes the function unique. Two functions agreeing at four points is nothing; two functions with the same derivative everywhere and the same value once are identical, and that uniqueness is the reason $e^x$ is the answer rather than an answer.

The bracketing argument has a similar gap. Two pictures show tangents that undershoot at base 2 and overshoot at base 3, and the conclusion drawn from them — that some base in between does neither — needs a fact neither picture contains: that the proportionality constant varies continuously with the base. Without that, a quantity below 1 in one drawing and above 1 in another could simply jump over the value, the way $1/x$ passes from negative to positive without ever being zero. The two frames are evidence that the answer is somewhere between 2 and 3. They are not, by themselves, evidence that it exists.

The picture also cannot show what happens off the real line. For complex arguments the exponential stops growing and starts rotating, which is the whole content of Euler’s formula — and a curve drawn on a page has no way to hint at it.

The ladder from here #

Rungs above: the natural logarithm as the area under $1/x$, which is a completely different-looking definition of the same constant. The differential equation $y' = ky$ and its solutions, which is where the physics enters. Half-lives and doubling times as one fact in two costumes. The limit definition and the series definition, reconciled by the binomial theorem — which brings in Pascal’s triangle. Compound interest with continuous withdrawal. Stirling’s approximation, where $e$ appears in factorials for no obvious reason. The derangement problem. Euler’s identity as a rotation. And $e^{-x^2}$, the shape of the bell curve, where the exponential turns up in a setting with no growth in it at all.

Where it turns up #

Because “rate of change proportional to current amount” is one of the most common situations in nature — as common as the circular motion behind every sine wave — $e$ appears wherever that description fits: radioactive decay, cooling objects, population growth without constraint, charge draining from a capacitor, drug concentrations. In every case the same differential equation, and therefore the same function.

It also appears in places with no growth at all. The normal distribution has $e^{-x^2/2}$ at its heart, which is where $e$ meets the bell curve that falls out of coin flips. And Euler’s formula, $e^{i\theta} = \cos\theta + i\sin\theta$, connects it to rotation — the subject of multiplying is turning — producing the identity $e^{i\pi} + 1 = 0$ that ties together the two constants nobody expected to meet.

The derangement problem is a good example of $e$ turning up where it has no business. Hand back $n$ hats at random; the probability that nobody gets their own hat approaches $1/e \approx 0.368$ as $n$ grows, and it converges so fast that the answer for $n = 6$ is already correct to three decimal places. There is no growth process, no continuous change, and no circle. There is just a factorial in the right place.

What is really being said #

The tangent-line picture is the definition that explains the others.

Compound interest converges to $e$ because continuous compounding is growth-proportional-to-size. The series sums to $e$ because term-by-term differentiation shifts it onto itself. The natural logarithm is natural because it inverts the function that needs no correction factor.

All three are consequences of one geometric property, which is why it is worth drawing. There is exactly one exponential curve that is its own slope, and the number that produces it had to be whatever it turned out to be.