$\pi$ from the unit circle, $\sqrt 2$ from the unit square but what about $e$? [duplicate]

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.)

Solution 1:

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 $1$.

Solution 2:

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.)

Solution 3:

It's the only number where this happens:

Solution 4:

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 $e$.

Solution 5:

$$e\approx 2.71828182846$$

Consider the equation:

$$f(n)=(1+\frac{1}{n})^n$$

As $n$ gets larger and larger, notice what the result approaches.

$$f(1)=2$$ $$f(2)=2.25$$ $$f(3)\approx2.3703703$$ $$...$$ $$f(100)\approx2.7048138$$ $$...$$