I was wondering how many definitions of exponential functions can we think of. The basic ones could be:

$$e^x:=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$ also $$e^x:=\lim_{n\to\infty}\bigg(1+\frac{x}{n}\bigg)^n$$ or this one: Define $e^x:\mathbb{R}\rightarrow\mathbb{R}\\$ as unique function satisfying: \begin{align} e^x\geq x+1\\ \forall x,y\in\mathbb{R}:e^{x+y}=e^xe^y \end{align} Can anyone come up with something unusual? (Possibly with some explanation or references).

16 Answers 16

The exponential function is the unique solution of the initial value problem

$y'(x)=y(x) , \quad y(0)=1$.

We can also define $e^x$ as follows:

  • the inverse function of $\ln x$, defining $\ln x$ independently as follows

$$\ln x := \int_1^x \frac{dt}{t}$$

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    Would an algebraic definition work, along the lines of: An exponential function is a homomorphism $exp \colon (\mathbb{R},+) \rightarrow (\mathbb{R},\cdot)$? Thus, $exp(r_1+r_2) = exp(r_1) \cdot exp(r_2)$ and $exp(0) = 1$ – j4nd3r53n Nov 7 at 15:47
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    And a good candidate for the definition of $\ln x$ would be $\ln x := \int_1^x \frac{dt}{t}$. – Daniel Schepler Nov 7 at 17:28
  • @DanielSchepler Thanks good point! That's commenti will be useful to the asker and future readers. – gimusi Nov 7 at 18:19
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    @j4nd3r53n It's not sufficient without an additional assumption such as the function being continuous at some point, or bounded on some interval, etc. See the comment below Paul Evans's answer. – Daniel Schepler Nov 7 at 18:23
  • @j4nd3r53n: that property is also shared by $a^x$. A careful analysis of the functional equation $f(x+y) =f(x) f(y) $ easentially leads to exponential function, but existence of such a function requires recourse to more traditional definitions. – Paramanand Singh Nov 8 at 9:08

Define the value at rationals via powers and roots and then show that there is a unique continuous function which agrees with these values.

First define it for the natural numbers:

Define $e^2 = e \times e$, $e^3 = e \times e \times e $, etc.

Now define it for other integers:

$e^0 = 1$, $e^{-n} = \frac{1}{e^n}$, etc.

Now for other rational numbers (getting a bit harder):

$e^{\frac{p}{q}} = \sqrt[q]{e^p}$

Finally for irrational numbers $x$, you will need to prove that this definition evaluated for any sequence of rational numbers which converge to $x$ has a limit and that it is the same for all sequences which converge to $x$.

This is hard, especially the last step, but I think that it fits a common naive idea of what exponentiation is. We usually learn it in this sequence.

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    Can you elaborate your answer with something concrete? – Michal Dvořák Nov 7 at 13:28
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    Good question. Either just right it down e.g. $e = 2.71828 . . .$ or use one of your first two definitions with $x = 1$. – badjohn Nov 7 at 15:43
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    You might just define $x^y$ at this point. – Ori Gurel-Gurevich Nov 7 at 16:04
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    True. My approach is probably more likely if you are trying to define exponentiation and you have not encountered $e$ yet. In that case, you probably won't be able to do the last stage very formally and will just assume it as obvious. I did not express it that way as the OP wanted definitions of $e^x$. – badjohn Nov 7 at 16:31
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    Yes, this approach is unusual but that is what was requested. – badjohn Nov 8 at 9:19

Throw $n$ balls into $n$ bins uniformly at random, and take $n \to \infty$. Define $\frac{1}{e}$ to be the limiting fraction of empty bins.

A vehicle moves from point $A$ to $B$ with speed always equal to the remaining distance to $B$. Define $1-\frac{1}{e}$ to be the fraction of distance covered after one unit of time.

Given positive $x$, consider a set of independent Bernoulli random variables with $\sum_{i=1}^n p_i = x$. As $n \to \infty$ and $\max_i p_i \to 0$, define $e^{-x}$ to be the probability that all are zero.

You can also define the exponential function like this: $$ e^x = \lim_{n\to\infty} \frac{f_n(x)}{f_n(-x)} $$ where $$ f_n(x) = \sum_{j=0}^n \frac{(2n-j)!}{j!(n-j)!}x^j $$

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    I have never seen this one, and it's interesting, could you please relate this somehow to the usual definitions? – Michal Dvořák Nov 8 at 7:15
  • @MichalDvořák - it's using Pade Approximants. The Pade approximant for $e^x$ for any orders (numerator and denominator) are well-known. I've just used the approximants for equal order numerator and denominator, which turns out to be relatively neat. Basically, for a given $n$, this formula will give the same Taylor expansion as $e^x$ for the first $2n$ terms. – Glen O Nov 11 at 9:43

Let $D_n$ be the number of permutations of $[n]$ without fixed points (i.e. derangements). Define $$ \frac{1}{e}=\lim_{n\to\infty}\frac{D_n}{n!}. $$

The exponential function is the unique smooth group isomorphism from the additive group of reals to the multiplicative group of positive reals whose derivative at zero is one. It is the Lie group exponential map of the latter group.

$e^x := \cos(-ix) + i \sin(-ix)$

(See Euler's formula)

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    Is it possible to define the sine and cosine of a complex number without using exponential? – Surb Nov 7 at 23:08
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    That's a good question! It seems like Euler's formula would be meaningless if it were not possible to define sin and cos for imaginary numbers. Perhaps the Taylor Series would make sufficient definitions? – Ben Jones Nov 7 at 23:47
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    Sine and cosine can be defined by the corresponding power series. But this seems a bit like running in circles... – weee Nov 7 at 23:48
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    So we keep running in circles and going "@weee..."? =) – user21820 Nov 8 at 8:04
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    @weee Where else would one run when dealing with sin and cos? :) – Hagen von Eitzen Nov 8 at 16:08

If you've defined $a^b$, then you can take any sequence with a limit involving $e$ and define $e^x$ in terms of that. For instance, $e^x=lim_{n\rightarrow \infty}\left[\frac{n!}{n^n\sqrt{(2n+\frac13\pi}}\right]^{\frac xn}$

EDIT: thanks to @HagenvonEitzen's comment.

I've often wondered if the following is sufficient for a general power function:

$$f(x+y) = f(x)f(y)$$

And then:

$$f(1) = e$$

for the base.

Think this is probably similar to @badjohn's answer.

EDIT: thanks to @CarstenS and @R

Turns out we must also demand that $f(x)$ is continuous or measurable.

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    Not without also demanding continuity. Take any additive $T$ with $T(1)=1$ and set $f(x) = \exp(T(x))$. – Carsten S Nov 7 at 16:44
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    @CarstenS: You can get by with weaker conditions than continuity. For example, measurability suffices (and, with additivity, of course implies continuity). – R.. Nov 8 at 2:28
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    @R.., yes I should have written "not without additional conditions". – Carsten S Nov 8 at 10:09
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    $f(-x)=\frac1{f(x)}$ is redundant as it follows from the first and the fact that $f$ is not identically zero – Hagen von Eitzen Nov 8 at 16:07

Similar to above. It pops up in applying Euler's Method for solving differential equations numerically:

Suppose we have $\frac{dy}{dx}=ky.$ And we have $y(0)=1$.

By Euler's Method: Pick some small $\Delta x$. Then Let: $$x_{n+1}=x_n+\Delta x$$ $$y_{n+1}=y_n+\frac{dy}{dx}\Delta x=y_n+(ky_n)\Delta x$$

With some rearranging: $$y_{n+1}=y_n(1+k\Delta x)\implies y_{n+p}=y_n(1+k\Delta x)^p$$ $$x_{n+p}=x_n+p\Delta x $$

Now we have that $x_0$=0 and $y_0=1$. So: $$x_{n+p}=(0)+p\Delta x$$ $$y_{n+p}=(1)(1+k\Delta x)^p$$

Suppose we know we want our last value of $x$ to b $z$ and we want $p$ steps. Then we want $\Delta x$ to be $z/p$.

Then letting $n=0$,

$$y_p=(1+k\frac{z}{p})^p$$

By the limite mentioned above, $y_p$ gets closer and closer to $e^{kz}$ as $p$ goes to infinity.

If we equip the one-dimensional manifold $(0,\infty)$ with the Riemannian metric $dx^2/x^2$, then the Riemannian exponential map at the point $1$ is the usual exponential function. This is the unique metric that makes the exponential function an exponential map.

Definition of $e$:

$$\lim_{h\rightarrow 0} \frac{e^h - 1}{h} = 1.$$

Define an exponent as a supremum of a set of a real number to rational powers.

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    This needs a bit more elaboration as to how you define arbitrary real exponents. – Paramanand Singh Nov 8 at 9:00
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    Also, this seems to define $e$ but not $x\mapsto e^x$ – Surb Nov 8 at 9:21

$f(x) = e^x > 0$ is such a function that

$$ \int\limits_{1}^{f(x)} \tfrac{1}{u} \, \mathrm{d}u \ = \ x$$

Similar things have been said, but not in that way:

$\exp$ is the only fixed point $f$ of the derivative (seen as function $D: C^{k+1}(\mathbb R) \to C^{k}(\mathbb R)$ for some $k \in [0,\infty]$, or between some other convenient spaces) which satisfies the normalization $f(0)=1$. (The set of all these fixed points is the one-dimensional eigenspace of $D$ for eigenvalue 1, and $\{\exp\}$ is a somewhat "canonical" basis for that space.)

Here's an "unusual" one: $e$ is the positive real number such that

$$ \sqrt{6\log\left(e\sqrt[4]{e}\sqrt[9]{e}\sqrt[16]{e}\sqrt[25]{e}\ldots\right)} = \pi. $$

  • Would you post some reference on this? This seems very interesting. – Michal Dvořák Nov 11 at 21:24
  • @MichalDvořák - it's just representing $\sum \frac1{n^2} = \frac{\pi^2}6$ in a really obtuse way. And it doesn't really define $e$ at all, since the logarithm has to be defined in terms of some base, and the $e$ just needs to match the base. – Glen O Nov 12 at 3:13
  • @GlenO Correct. I posted this as a "joke answer", so obviously this is not a definition. But does it deserve a downvote? – Flermat Nov 12 at 8:08

protected by Aloizio Macedo Nov 8 at 16:09

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