The Most Beautiful Formula in Mathematics (Part I)

A few years ago, while I was still in college, I gave a seminar entitled “The Most Beautiful Formula in Mathematics”. Many friends were unable to attend and asked for a transcript of the seminar, which I now provide.

The text is divided into two more or less independent parts, depending on your prior knowledge. In this first part we will talk about Euler’s number and see what makes it so special. In the second part (to be posted) we will study Euler’s formula itself, a beautiful connection between Euler’s number and trigonometric functions obtained with the help of complex numbers.

As a final note before we begin, I want to make it clear that most of the developments I will present were taken from other sources; however, due credit will be given.

I. Euler’s number

For a long time I was intrigued by the use of the number $e = 2.7182818...$ as the base of the so-called natural logarithm. How could this base be natural? Why this number in particular? How can an irrational number be more natural than any natural number? I found a satisfactory answer in Euler’s own work, presented masterfully in William Dunham’s book Euler: The Master of Us All.

Our starting point is exponentiation. More specifically, we will analyze powers of the form $a^{\omega}$ with $\omega$ “small”, that is, $\omega \rightarrow 0$ in the more modern language of limit theory. In Euler’s time, Differential and Integral Calculus had not yet gone through the formalization process that later became standard through the concept of limits, so Euler used the concept of infinitesimals: infinitely small quantities. In homage to Euler, I will avoid the use of limits in this article, using the concept of infinitesimal in an intuitive and non-rigorous way. Still following Euler, I will use the Greek letters $\psi$ and $\omega$ as symbols for these infinitesimal quantities.

Notice that $a^{\omega}$ will get closer and closer to $1$ as the value of $\omega$ approaches $0$. In other words, we can write

$$\begin{aligned} a^{\omega} = 1 + \psi \implies \psi = a^{\omega} - 1, \tag{1} \end{aligned}$$

with $\psi \rightarrow 0$ whenever $\omega \rightarrow 0$.

Let us take $a = 2$ and make $\omega$ smaller and smaller. We obtain the following table:

$\omega$	$2^\omega$	$\psi$	$k = \frac{\psi}{\omega}$
$0.1$	$1.071773463$	$0.071773463$	$0.717734625$
$0.01$	$1.006955550$	$0.006955550$	$0.695555006$
$0.001$	$1.000693387$	$0.000693387$	$0.693387463$
$0.0001$	$1.000069317$	$0.000069317$	$0.693171204$
$0.00001$	$1.000006931$	$0.000006931$	$0.693149583$
$0.000001$	$1.000000693$	$0.000000693$	$0.693147421$
$0.0000001$	$1.000000069$	$0.000000069$	$0.693147204$

Notice that as we decrease $\omega$, $\psi$ also decreases. If we define the proportion between the second and the first as $k = \frac{\psi}{\omega}$, we see that this proportion seems to approach a constant value, $k \approx 0.693$ (some will recognize this number!).

Let us see how the table looks for $a = 5$:

$\omega$	$5^\omega$	$\psi$	$k = \frac{\psi}{\omega}$
$0.1$	$1.174618943$	$0.174618943$	$1.746189431$
$0.01$	$1.016224591$	$0.016224591$	$1.622459127$
$0.001$	$1.001610734$	$0.001610734$	$1.610733753$
$0.0001$	$1.000160957$	$0.000160957$	$1.609567434$
$0.00001$	$1.000016095$	$0.000016095$	$1.609450864$
$0.000001$	$1.000001609$	$0.000001609$	$1.609439208$
$0.0000001$	$1.000000161$	$0.000000161$	$1.609438043$

The ratio $k$ is different for $a =5$, but it seems to approach a given value, just as in the case of $a = 2$.

Given that $k = \frac{\psi}{\omega} \implies \psi = k \omega$, we can rewrite equation $(1)$ as:

$$\begin{aligned} a^\omega = 1 + k \omega.\tag{2} \end{aligned}$$

From this equation, we can isolate $a$:

$$\begin{aligned} a = \left ( 1 + k \omega \right ) ^ {\frac{1}{\omega}}, \end{aligned}$$

where we extract the $\omega$ -th root of equation $(2)$.

To compute any power $x$ of $a$, not necessarily infinitesimal, we can write

$$\begin{aligned} a^x = \left ( 1 + k \omega \right ) ^ {\frac{x}{\omega}}. \end{aligned}$$

Notice that $j = \frac{x}{\omega} \rightarrow \infty$, given the infinitesimal nature of $\omega$ (a finite number divided by a quantity sufficiently close to zero will result in a number as large as we want). We can also write $\omega = \frac{x}{j}$ and substitute it into the equation above, together with the definition of $j$, which gives us

$$\begin{aligned} a^x = \left ( 1 + \frac{kx}{j} \right ) ^ j. \tag{3} \end{aligned}$$

Our intention is to apply the Binomial Theorem to this last equation. However, we will use a version with simplified binomial coefficients (canceling much of the factorials) which, incidentally, remains valid for the generalization of the binomial formula with real exponents first demonstrated by Newton. The Binomial Theorem tells us that, given $n \in \mathbb{N}$ and $x,y \in \mathbb{R}$:

$$\begin{aligned} (x+y)^n &= {n \choose 0 } x^n + {n \choose 1 }x^{n-1}y^1 + {n \choose 2 }x^{n-2}y^2 + {n \choose 3 }x^{n-3}y^3 + \cdots \\ &= \frac{n!}{0!n!}x^n + \frac{n!}{1!(n-1)!}x^{n-1}y^1 + \frac{n!}{2!(n-2)!}x^{n-2}y^2 + \frac{n!}{3!(n-3)}x^{n-3}y^3 + \cdots \\ &= 1x^n + \frac{n(n-1)!}{1!(n-1)!}x^{n-1}y^1 + \frac{n(n-1)(n-2)!}{2!(n-2)!}x^{n-2}y^2 + \frac{n(n-1)(n-2)(n-3)!}{3!(n-3)!}x^{n-3}y^3 + \cdots \\ &= x^n + \frac{n}{1!}x^{n-1}y^1 + \frac{n(n-1)}{2!}x^{n-2}y^2 + \frac{n(n-1)(n-2)}{3!}x^{n-3}y^3 + \cdots \end{aligned}$$

Applying the Binomial Theorem in the form above with $n = j$, $x = 1$ and $y = \frac{kx}{j}$, we get:

$$\begin{aligned} a^x &= 1 + \frac{j}{1!} \cdot \frac{kx}{j} + \frac{j(j-1)}{2!} \cdot \frac{k^2x^2}{j^2} + \frac{j(j-1)(j-2)}{3!} \cdot \frac{k^3x^3}{j^3} \\ &\quad + \frac{j(j-1)(j-2)(j-3)}{4!} \cdot \frac{k^4x^4}{j^4} + \dots \\ &= 1 + \frac{j}{j} \cdot \frac{kx}{1!}+ \frac{j(j-1)}{j^2} \cdot \frac{k^2x^2}{2!}+ \frac{j(j-1)(j-2)}{j^3} \cdot \frac{k^3x^3}{3!} \\ &\quad + \frac{j(j-1)(j-2)(j-3)}{j^4} \cdot \frac{k^4x^4}{4!} + \dots \\ &= 1 + \frac{j}{j} \cdot \frac{kx}{1!}+ \frac{j}{j} \left ( \frac{j-1}{j} \right ) \cdot \frac{k^2x^2}{2!}+ \\ &\quad \frac{j}{j} \left ( \frac{j-1}{j} \right ) \left ( \frac{j-2}{j} \right ) \cdot \frac{k^3x^3}{3!} \\ &\quad +\frac{j}{j} \left ( \frac{j-1}{j} \right ) \left ( \frac{j-2}{j} \right ) \left ( \frac{j-3}{j} \right ) \cdot \frac{k^4x^4}{4!} + \dots \\ &= 1 + 1 \cdot \frac{kx}{1!}+ 1 \left ( 1- \frac{1}{j} \right ) \cdot \frac{k^2x^2}{2!}+ \\ &\quad 1 \left ( 1- \frac{1}{j} \right ) \left (1 - \frac{2}{j} \right ) \cdot \frac{k^3x^3}{3!} \\ &\quad + 1 \left ( 1 - \frac{1}{j} \right ) \left ( 1- \frac{2}{j} \right ) \left ( 1- \frac{3}{j} \right ) \cdot \frac{k^4x^4}{4!} + \dots \end{aligned}$$

Now, if we remember that $j \rightarrow \infty$, we will see that the terms $\frac{1}{j}$, $\frac{2}{j}$, $\frac{3}{j}$, $\frac{4}{j}, \dots \rightarrow 0$, and the monstrosity above reduces to:

$$a^x = 1 + kx + \frac{k^2x^2}{2!} + \frac{k^3x^3}{3!} + \frac{k^4x^4}{4!} + \dots \tag{4}$$

Setting $x = 1$, we immediately obtain the value of $a$ in terms of $k$ alone:

$$a = 1 + k+ \frac{k^2}{2!} + \frac{k^3}{3!} + \frac{k^4}{4!} + \dots \tag{5}$$

As an example, taking $k = 0.693$ and using the first four terms of this equation, we obtain $a = 1.9982$, a value very close to $2$.

Among all possible values of $k$, which one would be the most “natural”? Which value of $k$ would make the calculation of $a^x$ easiest? If we take $k = 0$, we get $a = 1$ and, of course, all powers will be easy to compute, but everything would be very boring! The next candidate would be $k = 1$, and here things get interesting! The value of $a$ corresponding to $k = 1$ even has a special letter to represent it: $e$. Euler’s number finally appears!

Let us see what consequences we can draw from the fact that $k = 1$ corresponds to Euler’s number, establishing its special character:

1. If we set $k = 1$ in equation $(5)$, we get:

$$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots \tag{6}$$

2. If we set $k = 1$ in equation $(2)$, we get:

$$e^\omega = 1 + \omega,$$

that is, it is very easy to compute $e^\omega$ for infinitesimal $\omega$.

3. If we set $k = 1$ in equation $(3)$, we get:

$$e^x = \left ( 1 + \frac{x}{j} \right ) ^ j, \quad j \to \infty.$$

Taking $x = 1$ and using the modern language of limits, we have:

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

This last formula is generally used as the definition of Euler’s number in calculus textbooks.

4. Finally, if we set $k = 1$ in equation $(4)$, we get:

$$\begin{aligned} e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \tag{7}, \end{aligned}$$

and we see that it is easy to compute $e^x$ even for non-infinitesimal $x$. In fact, more than that, we can estimate the error when we use $(3)$ for small but non-infinitesimal values of $\omega$.

I recommend that the reader use a scientific calculator, a spreadsheet, or even a programming language to obtain estimates for Euler’s number and its powers through the equations listed above. As an example, thanks to the factorials in $(6)$, it is enough to sum $7$ terms to obtain a value correct to the third decimal place:

$$e \approx 2.718.$$

A more precise value would be

$$e = 2.718281828459045235360287471352662497757247\dots$$

If you would like to hear the first $5000$ digits of $e$, I made a video:

Relation to powers in other bases

Compare equation $(7)$ with $(4)$. The right-hand sides are practically the same! We just need to replace $x$ by $k$ in equation $(7)$, that is:

$$e^k = 1 + k + \frac{k^2}{2!} + \frac{k^3}{3!} + \frac{k^4}{4!} + \dots = a.$$

Then we can find $a$ in terms of $k$ through $e$:

$$a = e^k,$$

and thus we see that $k$ is nothing other than the natural logarithm of $a$:

$$k = \ln{a} \tag{8}.$$

The limiting values of $k$ obtained in the tables above are recognized as $\ln(2) = 0.6931471805599453...$ and $\ln(5) = 1.6094379124341003 ...$.

If you are not familiar with the logarithm function, observe that by definition it is precisely the function that gives us $k$ given $a$. It would be interesting if we had an easy way to compute it, would it not? After all, we quickly run into problems with the accuracy of the calculations if we use the table method above. We provide a series for computing the logarithm in the bonus section at the end, still following Euler.

Before that, however, we will leave a brief note on the question of convergence.

Note on infinite sums and convergence

We must emphasize that all these sums in $(4)$, $(5)$, $(6)$ and $(7)$ are infinite. We must always be careful when dealing with this type of sum, and we should observe under which conditions the sums make sense, that is, when (and if) they converge to a finite value. In this particular case, it is not hard to prove that the sums in question produce finite results for any $k, x \in \mathbb{R}$, thanks to the factorials in the denominators, which grow faster than any power present in the numerators. All of this can be demonstrated rigorously, but it would take even more time in this brief seminar. For now, contemplate $(5)$ written as follows and notice how quickly the denominators grow:

$$\begin{aligned} e^x &= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720} + \frac{x^7}{5040} \\ &\quad + \frac{x^8}{40320} + \frac{x^9}{362880} + \frac{x^{10}}{3628800} + \frac{x^{11}}{39916800} + \dots \end{aligned}$$

If we take $x = 2$, for example, the 11th-power term contributes only $\frac{2^{11}}{39916800} = \frac{2048}{39916800} \approx 0.0000513$ to the final result.

Bonus: A series for the natural logarithm function

Let us begin by isolating $k$ in $(2)$:

$$k = \frac{a^\omega - 1}{\omega}$$

To apply Newton’s binomial trick once more, let us set $a = 1 + x$ and remember that, by $(8)$, $k = \ln(a) = \ln(1 + x)$, so the equation above becomes

$$k = \ln(1 + x) = \frac{(1+x)^\omega - 1}{\omega} \tag{9}.$$

Now, by Newton’s binomial theorem, we have:

$$\begin{aligned} (1+x)^\omega &= 1 + \frac{\omega}{1!} x + \frac{\omega(\omega-1)}{2!} x^2 + \frac{\omega(\omega-1)(\omega -2)}{3!} x^3 \\ &\quad + \frac{\omega(\omega-1)(\omega-2)(\omega-3)}{4!} x^4 + \dots \end{aligned}$$

Passing the number $1$ to the left-hand side and dividing both sides by $\omega$, we get:

$$\begin{aligned} \frac{(1+x)^\omega -1}{\omega} &= \frac{1}{1!} x + \frac{(\omega-1)}{2!} x^2 + \frac{(\omega-1)(\omega -2)}{3!} x^3 \\ &\quad + \frac{(\omega-1)(\omega-2)(\omega-3)}{4!} x^4 + \dots \end{aligned}$$

On the other hand, $\omega$ is an infinitesimal, that is, $\omega \to 0$, so:

$$\begin{aligned} \ln(1+x) &= \frac{(1+x)^\omega -1}{\omega} \\ &= \frac{1}{1!}x + \frac{(0-1)}{2!}x^2 + \frac{(0-1)(0-2)}{3!}x^3 \\ &\quad + \frac{(0-1)(0-2)(0-3)}{4!}x^4 + \dots \\ &= \frac{1}{1!}x + \frac{-1}{2}x^2 + \frac{(-1)(-2)}{1\cdot2\cdot3}x^3 \\ &\quad + \frac{(-1)(-2)(-3)}{1\cdot2\cdot3\cdot4}x^4 + \dots \\ &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \end{aligned}$$

In other words, we have:

$$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \dots \tag{10}$$

Observe that here we no longer have factorials in the denominators and, in fact, this series does not converge for every $x$, but only for $-1 < x \leq 1$. Take $x = 2$, for example, and see how the partial sums oscillate and seem to grow larger and larger in magnitude:

$$\begin{aligned} s_1 &= 2 \\ s_2 &= 2 - \frac{2^2}{2} = 0 \\ s_3 &= 2 - \frac{2^2}{2} + \frac{2^3}{3} = 2.666... \\ s_4 &= 2 - \frac{2^2}{2} + \frac{2^3}{3} - \frac{2^4}{4} = -1.333... \\ s_5 &= 2 - \frac{2^2}{2} + \frac{2^3}{3} - \frac{2^4}{4} + \frac{2^5}{5} = 5.066... \\ s_6 &= 2 - \frac{2^2}{2} + \frac{2^3}{3} - \frac{2^4}{4} + \frac{2^5}{5} - \frac{2^6}{6} = 5.6 \\ s_7 &= 2 - \frac{2^2}{2} + \frac{2^3}{3} - \frac{2^4}{4} + \frac{2^5}{5} - \frac{2^6}{6} + \frac{2^7}{7} = 12.6857142857... \\ s_8 &= 2 - \frac{2^2}{2} + \frac{2^3}{3} - \frac{2^4}{4} + \frac{2^5}{5} - \frac{2^6}{6} + \frac{2^7}{7} - \frac{2^8}{8} = -19.3142857143... \end{aligned}$$

The correct value is $\ln(1 + 2) = \ln(3) = 1.0986122886681098...$.

Here, Euler once again sheds light on the problem, because by replacing $x$ with $-x$ in $(10)$, we have:

$$\ln(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \frac{x^6}{6} - \dots,$$

and, using a well-known property of logarithms:

$$\begin{aligned} \ln(1 + x) - \ln(1 - x)&= \ln \left( \frac{1 + x}{1 - x} \right) \\ &= \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \right) \\ &\quad - \left( -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} + \dots \right) \\ &= 2 \left(x + \frac{x^3}{3} + \frac{x^5}{5} + \dots \right) \end{aligned}$$

Let us highlight this last result:

$$\ln \left( \frac{1 + x}{1 - x} \right) = 2 \left(x + \frac{x^3}{3} + \frac{x^5}{5} + \dots \right)$$

Setting $x = \frac{1}{2}$, we have $\frac{1 + x}{1 - x} = \frac{1 + 1/2}{1 - 1/2} = 3$ and, from the last series obtained:

$$\begin{aligned} \ln(3) = \ln \left( \frac{1 + 1/2}{1 - 1/2} \right) &= 2 \left(\frac{1}{2} + \frac{1}{2^3 3} + \frac{1}{2^5 5} + \frac{1}{2^7 7} \dots \right) \\ &= 1 + \frac{1}{12} + \frac{1}{80} + \frac{1}{448} + \dots, \end{aligned}$$

and now the series converges to the correct value of $\ln(3)$. Summing only the first four terms shown above already gives us $\ln(3) \approx 1.0981$, correct to the third decimal place.