$\pi$ from the unit circle, $\sqrt 2$ from the unit square but what about $e$? If one wants to introduce $\pi$ to a not mathematically savvy person, the unit circle would be a good choice. The unit square would be the way to go for $\sqrt 2$. But what about $e$? I've reviewed the alternative characterizations on wikipedia and they all involve limiting processes in one form or another. So I was wondering: Is there a visual, easy-to-grasp way to present $e$? (Even if the presentation wouldn't be mathematically equivalent to the usual ones.)
 A: It's the only number where this happens:

A: The notion of period, which is introduced by Kontsevich and Zagier, would partially give a negative answer to your question. According to this article, it is now known whether $e$ is a period or not, though it is conjecturedly not a peroid. In particular, $e$ seems not to arise as an area or a length of a geometric figure defined by an algebraic equation.
This may imply that any description involving $e$ would eventually require some extent of advanced calculus.
We may circumvent this problem by arguing that the concept of $e$ arises rather naturally by considering some real-world problems. For example, we can think of compound interest:
Given an annual interest $100r$% with compounding period $1/n$ year, the relative value at the end of a year is
$$ \left( 1 + \frac{r}{n} \right)^{n}, $$
which converges to $e^{r}$ as $n \to \infty$ (that is, the period of interest goes to zero).
Here is another, though odd, characterization of $e$:
Suppose that a drunk man starts at $x = 0$ at time $t = 0$, and then at each second he walks to the right with a step size randomly and evenly between $0$ and $1$. That is, if $X_i$ are i.i.d. random variables having a uniform distribution on $[0, 1]$, then the position of the drunk man at time $t = n$ is $x = X_1 + \cdots + X_n$. Let $T$ be the time when his position first exceeds the point $x = 1$. Then the mean of $T$ is exactly $e$. 
A: 
$$e\approx 2.71828182846$$

Consider the equation:
$$f(n)=(1+\frac{1}{n})^n$$
As $n$ gets larger and larger, notice what the result approaches.
$$f(1)=2$$
$$f(2)=2.25$$
$$f(3)\approx2.3703703$$
$$...$$
$$f(100)\approx2.7048138$$
$$...$$
A: This one is not visual or graphical, but may be easiest to understand for a non-mathematician.
Suppose you have $\$1000$ and you want to put it in a bank account. You have picked a bank that, besides giving you an absurd interest over your money, gives you a choice between several interest schemes:


*

*An annual interest of $100\%$.

*An interest of $50\%$, but then twice a year.

*A quarterly interest of $25\%$.

*A monthly interest of $8.33\%$.

*...

*A daily interest of $0.27397\ldots\%$

*...
As you may have picked up, the interest is $1/k$, where $k$ is the number of times a year you get that interest. Of course, the interest is computed over the current amount of money in your bank account, including previous interest.
So which scheme would you choose? And how much money would you have after $1$ year?
A: The number $e$ is the number such that the area enclosed by the region bounded by $x=1$ on the left-hand side, the $X$ axis from below, $y=1/x$ from above and $x=e$ on the right-hand side is $1$. Hence, $e$ can be defined as the length (you need to add $1$, since you measure the length from $1$) you need to move along $X$ axis such that the area of the above enclosed figure is $1$.

A: A different take on the 'classical' limit that I think is my favorite way of thinking about $e$ recreationally (and a remarkably useful approximation for many games): "I take a six-sided die and roll it six times.  What are the odds I never roll '1' in those six rolls?  Okay, now I take a twenty-sided die and roll it twenty times.  What are the odds I never roll '1' in those 20 rolls?"  The latter value, of course, is $\left(\dfrac{19}{20}\right)^{20} = \left(1-\dfrac{1}{20}\right)^{20}\approx e^{-1}\approx .37$; in fact, it's within a 3% relative approximation.
This is a nice way of showing the limiting process in action 'in reality', especially for any gamers, and the way that the result (power-law) scales with number of rolls is an excellent back-of-the-envelope estimate for figuring probabilities on the fly; I can know that if I roll that d20 ten times, I've got roughly a $\sqrt{.36}\approx 60\%$ chance of not hitting a 20, or equivalently a $40\%$ chance of getting one.  (Or, to pick a different example, I can know that with 3 Plains left in my Magic sealed deck's 30-card library, my odds of drawing one in the next ten turns are only about 2/3rds.)
