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).
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:
$$\ln x := \int_1^x \frac{dt}{t}$$
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.
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.
You can also define the exponential function like this: $$ e^x = \lim_{n\to\infty} \frac{f_n(x)}{f_n(-x)} $$ where $$ f_n(x) = \sum_{j=0}^n \frac{(2n-j)!}{j!(n-j)!}x^j $$
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.
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.
$e^x := \cos(-ix) + i \sin(-ix)$
(See Euler's formula)
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, $$e^x= \lim_{n\rightarrow \infty} \left[ \frac{n!}{n^n\sqrt{(2n+\frac13)\pi}} \right] ^{x/n} $$
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.
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.
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.
$f(x) = e^x > 0$ is such a function that
$$ \int\limits_{1}^{f(x)} \tfrac{1}{u} \, \mathrm{d}u \ = \ x$$
Here is a very simple geometric construction which produces $f(x)=e^x$.
Take two objects on a flat level surface, one at the point $(0,0)$ and the other at $(0,1)$, that are connected by a piece of string. Slide the top object along the horizontal line $y=1$. The horizontal displacement of this object is given by $x$, and the trajectory of the bottom object (the path of least resistance) is called a tractix.
After sliding $x$ units horizontally, draw the line segment of length $1$ which is parallel to the string and which contains $(0,0)$ as one of its endpoints. The second point of this line segment lies on the unit circle. Draw a line through the circle endpoint and the point $(1,0)$. The $y$-intercept $b$ of this line is equal to $e^x$, and the point $(x,b)$ belongs to the graph of the exponential function $f$.
Here is an animation I made using GeoGebra: https://youtu.be/zzvwGl9WpX8
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
$$ \sqrt{6\log\left(e\sqrt[4]{e}\sqrt[9]{e}\sqrt[16]{e}\sqrt[25]{e}\ldots\right)} = \pi. $$
