It seems as though, in my analysis and calculus courses, in particular, a common cop-out when asked to prove an identity involving $e$, is the phrase "it's true by definition".

So, I'm trying to find as many definitions of $e$ in order to see just how many of these identities can actually be a definition of $e$.

So far, I've got the following (which are the ones most mathematicians know):

  • $e:=\lim\limits_{n \to \infty}(1+\frac{1}{n})^n=\lim\limits_{h \to 0}(1+h)^{1/h}$
  • $e:=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$
  • $e$ is the global maximum of the function $x^{1/x}$
  • $e$ is the real number satisfying $\int\limits_{1}^{e}\frac{1}{x}dx=1 \iff \begin{cases} \frac{d}{dx}[e^x]=e^x \\ \\ e^0=1 \end{cases} $

Does anyone have any more to add to the list?


  • $\begingroup$ @barakmanos Already got it! $\endgroup$
    – beep-boop
    Jun 14, 2014 at 14:36
  • $\begingroup$ I think most mathematicians consider the first one to be a theorem; it's only presented as a definition to beginners. Likewise, the third one is certainly not a definition. If you're looking for a list of "all identities involving $e$, then this is an impossibly broad and sort of useless question. If you're looking for just the accepted definitions, I think two and four are it. This amounts to "how many ways are there to define the exponential function", since this is the theoretical perspective on where $e$ comes from. $\endgroup$
    – Ryan Reich
    Jun 14, 2014 at 14:37
  • 6
    $\begingroup$ $e = \lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}}$ $\endgroup$ Jun 14, 2014 at 14:38
  • 2
    $\begingroup$ In fact, here is all of them: en.wikipedia.org/wiki/Representations_of_e $\endgroup$ Jun 14, 2014 at 14:39
  • 3
    $\begingroup$ @Ryan Reich: I am one of the mathematicians who thinks that the first definition in the list ( at least the $\lim_{n \to \infty}$ version) is a perfectly good way to define $e$. I agree about the real number version: I'm not sure how you define $(1+h)^{1/h}$ if $1/h \not \in \mathbb{N}$ if you don't already have some sort of $\log$ and exponential functions available. $\endgroup$ Jun 14, 2014 at 15:17

3 Answers 3


You really should be looking for definitions of the exponential function $e^x$, not definitions of $e$. Here are the most important ones that come to mind:

  1. $e^x$ is the unique function $f(x)$ satisfying $f'(x) = f(x)$ and $f(0) = 1$.
  2. $e^x$ is the inverse of the function $\displaystyle \ln x = \int_1^x \frac{dt}{t}$.
  3. $e^x$ is the power series $\displaystyle \sum_{n \ge 0} \frac{x^n}{n!}$.
  4. $e^x$ is the limit $\displaystyle \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$.

I say this because once you start thinking about $e^x$, which is by far the more fundamental object, and not $e$, the relationship between the definitions becomes much more transparent. Here are short sketches of proofs that the definitions above are all equivalent:

$1 \Leftrightarrow 2$: if $\frac{d}{dx} g(x) = \frac{1}{x}$ then

$$\frac{d}{dx} g^{-1}(x) = \frac{1}{g'(g^{-1}(x))} = g^{-1}(x).$$

Conversely, if $\frac{d}{dx} g(x) = g(x)$ then

$$\frac{d}{dx} g^{-1}(x) = \frac{1}{g'(g^{-1}(x))} = \frac{1}{g(g^{-1}(x))} = \frac{1}{x}.$$

$1 \Leftrightarrow 3$: if $f'(x) = f(x)$ then $f^{(n)}(x) = f(x)$ for all $n$, hence $f^{(n)}(0) = f(0) = 1$, so the Taylor series of $f(x)$ has all coefficients equal to $1$. Conversely, the function with that Taylor series is its own derivative and satisfies $f'(0) = 1$ using the fact that power series are term-by-term differentiable inside their interval of convergence.

$1 \Leftrightarrow 4$: $\displaystyle \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$ is the result of using Euler's method to compute $f(x)$, where $f$ satisfies $f'(t) = f(t)$ and $f(0) = 1$, as the step size $n$ goes to $\infty$.

(My personal opinion is that the first definition is the most fundamental one; in general uniqueness statements are very powerful. For example, there is a very short proof using the first definition, which I invite you to find, that $e^{x + y} = e^x e^y$.)

  • $\begingroup$ 5. $e$ is the unique real number that makes the function $e^x$ (defined by extending $e^{a/b}$ by continuity) satisfy property (1). $\endgroup$
    – Jack M
    Jun 15, 2014 at 7:28
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    $\begingroup$ Also, as an extra 2 cents, once you have the exponential function, I find a more natural definition of $e$ is as the constant ratio $f(x+1)/f(x)$, rather than as $f(1)$, which makes it seem a little arbitrary ($f(1)$ exists for all functions $f$, but the fact that that ratio is constant is a remarkable property of the exponential function). $\endgroup$
    – Jack M
    Jun 15, 2014 at 7:31
  • $\begingroup$ I would say that we shouldn't define the exponential function by the differential equation, because you would need to prove that it exists, which is most easily done by the definition via the power series. I would however agree that the differential equation is the one that lies at the heart of the exponential function. Also, the power series works for complex numbers, while the second definition would require a line integral in the complex plane and Cauchy-Goursat theorem or something. $\endgroup$
    – user21820
    Dec 4, 2014 at 5:22
  • $\begingroup$ @user21820: I think it's important to distinguish between definitions and constructions. I agree that it's very convenient to construct $e^x$ via power series, but the fundamental reason why you're looking at that power series and not some other power series is because it solves a differential equation that you care about. I am fine with definitions including theorems that the definitions are well-defined. $\endgroup$ Dec 4, 2014 at 5:46
  • 1
    $\begingroup$ @athos: $e^{x+y}$ and $e^x e^y$ are both solutions to the ODE $f'(x) = f(x)$ with initial condition $f(0) = e^y$. So by the uniqueness theorem they must be equal. $\endgroup$ Sep 29, 2022 at 17:51

$$e=\lim_{n\to\infty}\sqrt[\large^n]{\text{LCM}[1,2,3,\ldots,n]},$$ where LCM stands for least common multiple.

  • $\begingroup$ What's up with the floor sign? $\endgroup$ Jun 15, 2014 at 4:45
  • $\begingroup$ It's not a floor sign. The square brackets represent, among other things, the LCM, and the round ones the GCD. Though such conventions may vary by country. $\endgroup$
    – Lucian
    Jun 15, 2014 at 5:36
  • $\begingroup$ Huh. The tops of the brackets aren't rendering at default zoom level for me; they only show up at 110%. $\endgroup$ Jun 15, 2014 at 5:46
  • $\begingroup$ Same for me @user2357112. $\endgroup$
    – Kaj Hansen
    Jun 15, 2014 at 5:53
  • $\begingroup$ Is it better now? Let me guess: You were using FireFox? $\endgroup$
    – Lucian
    Jun 15, 2014 at 6:38

Actualy for certain numbers like $e$ and $\pi$ an almost enormous amount of defining relations or identities can be produced.

Think of the case of $\pi$, by using trigonometric identities (or even number-theoretic identities) many many defining relations for $\pi$ can be produced (i think there is also a systematic procedure to produce new defining relations).

A very close case is for $e$, due to being related to similar hyperbolic trigonometric functions (except the purely exponential-analytic identities), a very large amount of identities which can be used as definitions or representations can be found.

Some of them can be found online as in here, however they do not exhaust all possible identities that can serve as definitions.

Finally an identity that relates $i$, $\pi$, $e$, $0$ and $1$ (which was a favorite of R. Feynman) is the Euler identity:



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