4
$\begingroup$

At the beginning of a calculus course, we encounter two famous limits. They are $$\lim_{x\to0}\frac{\sin x}{x}\qquad\text{and}\qquad\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n.$$ I'm not sure if Cauchy was the first person in history to calculate the limit $\lim\limits_{x\to0}\dfrac{\sin x}{x}$. On page 64 of his famous book: Cours d'Analysis, he calculated this limit. I want to know who was the first person to calculate the limit $\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n$?

$\endgroup$
1
  • 5
    $\begingroup$ Jacob Bernoulli was the first to propose the problem of continuous compounding interest where the limit naturally arises. According to Boyer's A History of Mathematics, Bernoulli showed that the limit exists and is less than three, though it is not made clear how many decimal places he may have calculated. $\endgroup$
    – nwr
    Aug 8, 2023 at 4:02

2 Answers 2

5
$\begingroup$

First of all, what does it exactly mean to "calculate" this limit? The limit equals to $e$, but this is the definition of $e$. So one only has to prove that the limit exists, and then denote it by some letter. The fact that this limit exists was known to Napier, who calculated it with at least 12 significant digits. (Though Napier had neither a definition of a limit nor definition of a real number). The formal definition of the number $e$ is due to Euler, and it was called the "Euler constant" for this reason.

See also Which came first, the natural logarithm or the base of the natural logarithm?

$\endgroup$
3
5
$\begingroup$

I followed up on a pointer given by @nwr in comments that Jacob Bernoulli first encountered $e$ in the context of computing continuous compound interest. Sure enough, the article "The number $e$" at MacTutor states:

Perhaps surprisingly, since this work on logarithms had come so close to recognising the number $e$, when $e$ is first "discovered" it is not through the notion of logarithm at all but rather through a study of compound interest. In 1683 Jacob Bernoulli looked at the problem of compound interest and, in examining continuous compound interest, he tried to find the limit of $\left(1+\frac{1}{n}\right)^{n}$ as $n$ tends to infinity. He used the binomial theorem to show that the limit had to lie between $2$ and $3$ so we could consider this to be the first approximation found to $e$.

The article cites Eli Maor, $e$ : the story of a number, Princeton: Princeton University Press 1994 and J. L. Coolidge, "The number $e$." The American Mathematical Monthly, Vol. 57, No. 9, Nov. 1950, pp. 591-602 (scan), but neither publication cites a relevant publication by Bernoulli. Since there are other publications that mention work from 1683 on compound interest by Jacob Bernoulli when recounting the history of $e$, I spent a couple of hours trying to find that, but to no avail. I am now convinced that MacTutor is in error. At long last, I consulted M. Cantor's work on the history of mathematics:

Moritz Cantor, Vorlesungen über Geschichte der Mathematik, Vol. 3. Leizig: B.G. Teubner 1898.

Starting at page 51 and continuing onto page 52, Cantor describes Jacob Bernoulli's computation of continuous compound interest and provides the relevant source from his collected works:

No. XL. Jacobi Bernoulli, "Quæstiones Nonnullaæ de Usuris, Cum solutione Problematis de sorte Aleatorum, propositi in Ephemerid. Gallic. A. 1685. Art. 25." Jacobi Bernoulli, Basileensis, Opera. Vol. 1, Geneva: Cramer & Philibert 1744, pp. 427-431 (scan)

The original publication reproduced in Opera is from 1690 and gives a solution for a problem posed in 1685:

J. B., "Quæstiones nonnullæ de usuris, cum solutione Problematis de Sorte Alearum, propositi in Ephem. Gall. A. 1685, artic. 25.", Acta Eruditorum Lipsiæ, May 1690, pp. 219-223 (scan)

The statement relevant to the asker's question appears on page 222:

si $a=b$, debebitur plusquam $2\frac{1}{2}a$, & minus quam $3a$

which is equivalent to the statement that $2\frac{1}{2}<e<3 $. These values derive in straightforward manner from a series expansion given earlier on the same page,

$$a + b + \frac{b^2}{2a} + \frac{b^3}{2\cdot3a^2}+ \ldots = a\left(1 +\frac{b}{a} +\frac{1}{1\cdot 2}\left(\frac{b}{a}\right)^2 + \frac{1}{1\cdot 2\cdot 3}\left(\frac{b}{a}\right)^3 + \ldots\right)$$

so when $a=b$ as specified by the quoted text the first three terms of the expression in parentheses immediately provide the stated lower bound of $2\frac{1}{2}$. It is interesting to note that Bernoulli saw a use for the series in geometry, in particular as it relates to logarithmic curves.

$\endgroup$
1
  • 2
    $\begingroup$ Nicely researched. I think we can get the upper bound of $3$ by noting that $\frac{1}{n!} \lt \frac{1}{2^n}$ so when $a = b = 1$ the series is less than $1 + 1 + \frac12 + \frac{1}{2^2} + ...$ $\endgroup$
    – nwr
    Aug 10, 2023 at 4:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.