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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.)

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  • $\begingroup$ You could demonstrate that $e^x$ is invariant under derivation? Is that too mathematical? $\endgroup$
    – Ian Coley
    May 10, 2013 at 16:34
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    $\begingroup$ @FrankMcGovern, in comparison to the circle and square it's somewhat mathematical. I was looking for something more tangible. $\endgroup$
    – Leo
    May 10, 2013 at 16:37
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    $\begingroup$ The most "geometric" is that the area under the curve $(x,1/x)$ with $x\in[1,e)$ is $1$. That doesn't seem nearly as intuitive, however - it is not a length or area. $e$ comes up a lot as a limit - for example, the derangement problem, or when dealing with compound interest. $\endgroup$ May 10, 2013 at 16:40
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    $\begingroup$ At heart I think this is the same question as math.stackexchange.com/questions/159707/… which has been asked here a few times over. $\endgroup$
    – Erick Wong
    May 10, 2013 at 16:41
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    $\begingroup$ I think this question is rescuable as an independent question; the other question specifically asks for geometric ways of constructing $e$, whereas this question asks for intuitive ways of describing $e$ to a layperson, which certainly don't need to be geometric. $\endgroup$ May 10, 2013 at 18:01

6 Answers 6

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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 $1$.

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    $\begingroup$ I am a kid, I don't understand areas or integral. Now? $\endgroup$
    – Inceptio
    May 10, 2013 at 16:40
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    $\begingroup$ You don't need integrals to understand area. $\endgroup$
    – user17762
    May 10, 2013 at 16:43
  • $\begingroup$ Agreed..Was being sarcastic(+1) $\endgroup$
    – Inceptio
    May 10, 2013 at 16:50
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    $\begingroup$ @Inceptio. If you don't understand area, it's liquely you won't understand $pi$ and $\sqrt{2}$ either. $\endgroup$ May 10, 2013 at 17:12
  • $\begingroup$ @arbautjc: How will you bring in $\pi$ to a kid? With a pictorial representation. He/she understands the concept of $\pi$, and anyway, $e$ has not fallen from heaven, it can be understood even by a small kid if there existed a very elementary proof for this. $\endgroup$
    – Inceptio
    May 10, 2013 at 17:15
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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.)

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  • $\begingroup$ Similarly, if a lazy coat room attendant for a large event hands back coats randomly, what is the approximate probability that nobody gets their own coat back? (The `hat' variant is more commonly told but seems a bit contrived.) $\endgroup$
    – awwalker
    May 10, 2013 at 17:32
  • $\begingroup$ Thank you for the great answer. Even though I wouldn't try to use this way to explain $e$ to a person who doesn't understand limiting process (the odds are that he doesn't understand probability either, since it's more advanced in my opinion). Yet I highly enjoyed the answer myself. Do you have a nice example of a word problem converging to $e$ instead of $e^{-1}$? $\endgroup$
    – Leo
    May 13, 2013 at 2:40
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It's the only number where this happens:

enter image description here

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    $\begingroup$ What just happened? $\endgroup$
    – Pål GD
    May 10, 2013 at 18:04
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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 $e$.

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$$e\approx 2.71828182846$$

Consider the equation:

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

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

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

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    $\begingroup$ Have you read the question? $\endgroup$ May 10, 2013 at 17:21
  • $\begingroup$ @MartinBrandenburg Why yes, I did read the question I answered. $\endgroup$ May 10, 2013 at 17:38
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    $\begingroup$ What's visual and easy-to-grasp in this answer? $\endgroup$
    – Roland
    Mar 9, 2016 at 23:24
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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:

  1. An annual interest of $100\%$.

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

  3. A quarterly interest of $25\%$.

  4. A monthly interest of $8.33\%$.

  5. ...

  6. A daily interest of $0.27397\ldots\%$

  7. ...

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?

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