Euler's constant from geometry

Often math invents new operations—special functions—to solve geometry problems. For example, the Pythagorean theorem

\begin{align} \label{eq:pytha} a^{2} + b^{2} = c^{2} \end{align}

comes from the geometry of a right triangle with side lengths \(a, b, c\).

We define the square root function to solve for \(c\)

\begin{align} \label{eq:pythasqrt} c(a,b) = \sqrt{a^{2} + b^{2}}. \end{align}

Similarly the unit circle consists of all points \((x,y)\) with

\begin{align} x^{2} + y^{2} = 1. \end{align}

Solving for \(y(x)\)

\begin{align} \label{eq:circle} y(x) = \sqrt{1-x^{2}} \end{align}

we can now find the area of a circle by computing the area under the curve \(y(x)\) \eqref{eq:circle}

\begin{align} \label{eq:circlearea} A = 4\int_{0}^{1} \sqrt{1-x^{2}} \dd{x} = \frac{\arcsin(1)}{2} \end{align}

The integral in \eqref{eq:circlearea} is given in general by the special function \(\arcsin\), so we are led to a definition of the constant \(\pi\):

\begin{align} \label{eq:pidef} \pi := 2 \arcsin(1). \end{align}

The equation \eqref{eq:pidef} is exactly the definition that "\(\pi\) is the area of a circle with radius 1", expressed as a mathematical equation using the special function \(\arcsin\), which we view as an operation that numerically answers a geometric problem.

Let's try the same approach with a new curve: the hyperbola \(y(x) = 1/x\), defined for \(x > 0\). The area under the curve from \(1\) to some point \(x\) is given by

\begin{align} \label{eq:hyperbolaarea} A(x) = \int_{1}^{x} \frac{1}{t} \dd{t} \end{align}

We don't know how to evaluate this integral, but we can simply define it as a new special function

\begin{align} \label{eq:logint} \ln x := \int_{1}^{x} \frac{1}{t} \dd{t} \end{align}

called the logarithm. Even without computing it, we can see qualitatively that: the area is positive for \(x>1\), zero at \(x=1\), and strictly increases as \(x\) increases.

So there must be some number \(e > 0\) for which

\begin{align} \label{eq:euler} \ln e = 1. \end{align}

Let's define the inverse of the logarithm as the exponential \(\exp\)

\begin{align} e := \exp(1). \end{align}

Now we have an exact parallel to the definition of \(\pi\) in \eqref{eq:pidef}: both constants arise from integrals that compute geometric quantities (areas under curves) and both require special functions to express the answers.

So, we have a definition of \(e\) as it relates to the area under a hyperbola. At this point, \(e\) has nothing to do with exponentiation—but we should expect this. The fact that \(\pi\) is defined by circle geometry doesn't immediately connect to side length ratios of right triangles either. To make this connection, we have to actually study the trigonometric functions, not just define them. In our case, we study \(\exp\) and \(\ln\).

It turns out that logarithms have a surprising property: if we rescale the length dimension of the hyperbola in \eqref{eq:logint} by \(a\), the area scales as

\begin{align} \int_{x_{1}}^{x_{2}} \frac{1}{t} \dd{t} = \int^{a x_{2}}_{a x_{1}} \frac{1}{t}\dd{t} \end{align}

from which one can derive² the miraculous property

\begin{align} \ln(xy) = \ln x + \ln y. \end{align}

This identity—discovered by John Napier in 1614—is what made logarithms important in the first place: they turn multiplication into addition. This means that the inverse of the logarithm must turn addition into multiplication!

\begin{align} \label{eq:explaw} \exp(x + y) = \exp(x)\exp(y). \end{align}

With this property and the fact that the special functions \(\exp\) and \(\ln\) are defined for real numbers, we can generalize the definition of exponentiation beyond integers: for any base \(b>0\),

\begin{align} b^{x} := \exp(x \ln b). \end{align}

If \(x\) happens to be an integer, we recover the usual definition, for example if \(x = 3\)

\begin{align*} b^{3} &= \exp(3 \ln b) \\ &= \exp(\ln b + \ln b + \ln b) \\ &= \exp(\ln b) \cdot \exp(\ln b) \cdot \exp(\ln b) \\ &= b \cdot b \cdot b. \end{align*}

While we have defined \(\exp, \ln\) geometrically, we can derive the important property \eqref{eq:explaw} which connects \(e\) with the operation of exponentiation as repeated multiplication.

We can now answer the classical question: given an annual interest rate \(R\), what is the equivalent continuously compounded interest rate \(r\)? That is:

\begin{align} 1+R = \lim_{n \to \infty}\qty(1 + \frac{r}{n})^{n}. \end{align}

Taking logarithms,

\begin{align} \ln\qty[ \qty(1+\frac{r}{n})^{n}] = n \ln(1 + \frac{r}{n}). \end{align}

For large \(n\), \(r/n\) is small and we can estimate³

\begin{align} \ln(1 + \frac{r}{n}) \approx \frac{r}{n} \end{align}

and we have

\begin{align} \ln(1+R) = \lim_{n \to \infty} n \ln(1 + \frac{r}{n}) = r \end{align}

therefore taking \(\exp\) of both sides gives

\begin{align} 1 + R &= \exp(r) \end{align}

as we saw in the beginning with \eqref{eq:bernoulliformula}.

The conceptual and mathematical motivation for studying \(e\) depends on ideas from calculus. Still, it's possible to present the special functions \(\exp, \ln\) without it, in the same way trigonometric functions are presented, focusing on its algebraic properties like \eqref{eq:explaw}. Given their importance in e.g., defining pH in high school chemistry, deferring until a proper calculus course may be impractical.

But, at the very least we can emphasize \(\exp\) and \(\ln\) as proper functions, as we do with \(\sin, \cos, \tan\), instead of as a generalization of exponentiation as repeated multiplication. This latter fact is a property of the special function, not its definition.⁴ Then, near the end of the curriculum, students can be introduced to formulas like Bernoulli's or the differential equation

\begin{align} \label{eq:diffeq} \dv{x} f(x) = f(x) \end{align}

without the expectation that they should mathematically evaluate any infinite limits.

Math formulas are intimidating when divorced from the necessary context and intuition. For trigonometric functions, this intuition comes from the simple geometry of circles and right triangles. Unfortunately, for \(\exp, \ln\) this intuition comes only from the infinitesimal calculus of derivatives and integrals. Ideally, American math education should wrestle with this in an intelligent way.