Hardly a research level question, but interesting nonetheless, I hope. Pi is easy, but not e. Where could I start?
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7$\begingroup$ en.wikipedia.org/wiki/… $\endgroup$– Martin BrandenburgCommented Oct 9, 2010 at 10:48
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6$\begingroup$ Something that should have been asked first: where did the ten year old see $e$ that s/he's asking you about it? $\endgroup$– J. M. isn't a mathematicianCommented Oct 9, 2010 at 12:07
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6$\begingroup$ Why is this question still on MO? $\endgroup$– Daniel PomerleanoCommented Oct 9, 2010 at 14:14
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7$\begingroup$ I would recommend defining it as the cardinality of the groupoid of finite sets. $\endgroup$– Vivek ShendeCommented Oct 9, 2010 at 19:45
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6$\begingroup$ @JM: FWIW I told my kids about $\pi$ when the oldest was about 5, just as an example of a number that was interesting but not a whole number. I vividly remember the 3-year-old then insisting that we should count $1,2,3,\pi,4,5,\ldots$ after that :-) We got onto $e$ later on, when one of them asked me if there were any other numbers "like $\pi$" (by which they meant "not a whole number, but had a concise name"). Now my kids know $e=2.718\ldots$ but don't have a clue why $e$ is at all relevant to anything. To get back to the point though, I have no idea what I'll say when one of them asks! $\endgroup$– Kevin BuzzardCommented Oct 9, 2010 at 21:53
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12 Answers
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I learned about the "secretary problem" when I was about 10 years old from one of Martin Gardner's books. Though I thought is was cool and amazing, I don't think it gave me much insight into $e$.
Here's a way to introduce $e$ with only addition and multiplication, in the form of a game.
Tell him he's got a "budget" of say, 100 to work with, and his goal is to pick a bunch of (positive) numbers, not necessarily whole numbers, that add up to 100, where he tries to make the product as large as possible.
In his mind, he might first think to break his 100 as 50$\times$50, then realize that 25$\times$25$\times$25$\times$25 is even better, then 10$\times$10$\times$10$\times$10$\times$10$\times$10$\times$10$\times$10$\times$10$\times$10 is even better, and so on. The more numbers you split it into, the better!
But wait...
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1$\begingroup$ Could someone explain this? I can see with calculus that the optimum value of $(100/x)^x$ is at $x=100/e$ (corresponding to something like 100/e copies of e), but (1) how would a 10-year-old arrive close to this without calculus?, and (2) even if she sees (by trying out everything with a calculator, say) that the answer for 100 is $(100/37)^{37}$, how to get from there to an understanding of $e$ in general? $\endgroup$ Commented Oct 23, 2012 at 9:27
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$\begingroup$ Guys I learnt about $e$ when I was 8!!! $\endgroup$ Commented Dec 13, 2013 at 17:52
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If you can explain $\pi$, then you can also explain radians, and this leads to a geometric interpretation of $e$ that is not in the wikipedia article Martin Brandenburg referred to.
Get a big sheet of paper and make a polar coordinate grid, and ask your ten-year-old to put a pencil down somewhere away from the origin, and try to draw a continuous curve that meets each radial line in a 45 degree angle. The curve will be a logarithmic spiral that spirals in toward the origin. Next, have the child trace along the curve toward the origin, starting at any point and stopping after 1 radian's worth of angular measure has been traversed. How much further away from the origin is the starting point than the stopping point? The answer is: by a factor of $e$. (By similarity considerations, it doesn't matter where you start.)
You could also make a little story out of this by getting the child to imagine four ants placed on the corners of a square, each facing its neighbor in the clockwise direction when looking down at them, and then each chasing its neighbor all moving at the same speed. The trajectories they trace out are logarithmic spirals which tend toward the origin, and which cut radial lines at 45 degree angles. I'll leave remaining details of the story to you.
In conjunction with such experiments, you could examine spirals on seashells (which are approximately logarithmic spirals), and notice that similarity of the shape of the spiral no matter how you turn it. Again, the spirals cut a "radial line" invariably at the same angle. You can then perform thought experiments on the sea shell along the same lines as above.
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4$\begingroup$ I personally prefer the view of the logarithmic/equiangular spiral as the path a moth takes when it is flying towards a light bulb or a candle. $\endgroup$ Commented Oct 9, 2010 at 13:53
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$\begingroup$ I was not aware of that, J.M. -- thanks! $\endgroup$ Commented Oct 9, 2010 at 14:03
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Via percents in a bank? If we have a bank deposit for $x$ per annum, then we get from $N$ dollars $N(1+x)$ after a year. But if we get $x/12$ every month, we get $N(1+x/12)^{12}$ after a year. If we get $x/365$ every day, we get $N(1+x/365)^{365}$ after year, and if we get our percents continuously, we get $Ne^x$ after a year.
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If I were to take your title
"How do I explain the number e to a ten year old?"
literally,
then my answer would be: you don't.
Assume that you are in the (realistic!) situation of having a ten year old in front of you, and you wanting to tell him/her something about $e$.
Then it's probably better to try to serve the interests of the ten year old, and not those of the number $e$ (the number $e$ being a meme, it has its own interests). In other words, try to tell him/her something about math that s/he will understand, and will thus appreciate.
PS: My guess is that 15 years old is a good age for starting to learn about $e$. It is pointless to try to learn about $e$ before having acquired a deep intuitive understanding of the notion of exponential growth. And I would bet that that's assimilated only relatively late in the cognitive development of a child.
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1$\begingroup$ It's a YMMV thing, I think; that's why I asked that question of where the kid saw $e$ in the comments. $\endgroup$ Commented Oct 9, 2010 at 14:14
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1$\begingroup$ That sounds a little bit deterministic. I usually find the "I learned differential topology when I was six" anecdotes pretty annoying, but in Brazil it's quite common for exponentials and logarithms to be taught in 8th grade, i.e. 13 or 14 y.o. I think that means that an interested and well-taught child could easily understand it much earlier. You're quite right about exponential growth being a prerequisite, but I think the more mundane base 2 (think bacteria dividing) is understandable quite early on. $\endgroup$– PietroCommented Oct 9, 2010 at 19:54
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1$\begingroup$ Concerning "in Brazil it's quite common for exponentials and logarithms to be taught in 8th grade, i.e. 13 or 14 y.o.": My mother worked with Jean Piaget (see en.wikipedia.org/wiki/Jean_Piaget) and researched the way mathematics is taught in primary schools. According to her, there is a widespread tendency to try to teach mathematical concepts at an age where the majority of children are not able to assimilate them. Of course, children eventually understand those concepts... but the main reason for that, is simply that they grow up. $\endgroup$ Commented Oct 9, 2010 at 22:18
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$\begingroup$ I like all of the above problems that give a novice an idea of where $e$ might show up. Little stories or anecdotes like that interest the child in thinking, not just mathematics. In the end, regardless of what you tell the 10 year old, you have served their interest as long as you have kept their attention. $\endgroup$ Commented Oct 11, 2010 at 0:05
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There is a nice way that I learned from Martin Gardner's books when I was young(er).
Imagine the following situation. There is a party with $N$ people. All of them throw their hats in the middle of the room, and then each of them takes one hat randomly. What is the probability that nobody gets its own hat?
The surprising answer is that this probability equals
$1-1+\frac{1}{2!}-\frac{1}{3!} + \cdots + \frac{(-1)^N}{N!}$,
which goes to $\frac{1}{e}=0.36788...$ for $N \to \infty$.
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1$\begingroup$ This is the beautiful problem of derangements en.wikipedia.org/wiki/Derangement I can't see an answer to the OP's question here either (proving that you get 1/e requires the definition of e :) ). $\endgroup$– hceCommented Oct 9, 2010 at 13:05
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$\begingroup$ Well, no sorry of course you can also define 1/e as the limit of the probability you quote. I think it would still make for a mysterious explanation though. $\endgroup$– hceCommented Oct 9, 2010 at 13:13
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$\begingroup$ The technical term for this quantity is the number of derangements of N. en.wikipedia.org/wiki/Derangement is surprisingly unhelpful for such simple subject matter; still I learned that some people use the notation !n for the number of derangements of n. They also link to a note by John Baez that looks nice: math.ucr.edu/home/baez/qg-winter2004/derangement.pdf $\endgroup$ Commented Oct 9, 2010 at 13:13
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Here is one way which I learned from Clio Cresswell's Mathematics and Sex, although unfortunately I'm not sure how to prove it. Suppose you are sure that you will meet exactly $n$ suitable marriage partners in your life, but you don't know which one is the right one for you. As you meet each of your marriage partners in turn, you can choose to marry or reject them, and you cannot return to a marriage partner you have previously rejected. All you can do is maintain a list of how suitable each past marriage partner has been and compare that to your current suitor. What is the optimum strategy to maximize your marital bliss?
As it turns out, the answer (under these somewhat artificial conditions) is to reject the first $\frac{n}{e}$ of your partners, then marry the next partner after that which is as suitable or more suitable than your previous partners.
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3$\begingroup$ Sounds like this one: mathworld.wolfram.com/SultansDowryProblem.html $\endgroup$ Commented Oct 9, 2010 at 11:40
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$\begingroup$ Thanks! Yes, that's exactly the result I'm referring to. $\endgroup$ Commented Oct 9, 2010 at 11:45
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3$\begingroup$ I think this is best known as the secretary problem: en.wikipedia.org/wiki/Secretary_problem I frankly can't see an answer to the OP's question though. $\endgroup$– hceCommented Oct 9, 2010 at 13:01
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1$\begingroup$ I don't like the name "secretary problem" though. If you consider marriage, it is easy to imagine that you only want to marry the most suitable partners, whereas you might satisfy with the second best secretary. Also, it's quite natural to interview all secretary candidates first and decide about which one to employ only later, whereas you can't just do that with women. $\endgroup$ Commented Oct 11, 2010 at 10:46
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$\begingroup$ At least you still can try ;-) $\endgroup$ Commented Mar 4, 2012 at 23:14
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When I was this age, I loved the factorials. So why not try to explain it via the Stirling formula? That $n!$ is ``rather close'' to $n^n$ but you need to adjust it a bit via division by $e^n$.
Also, if he knows logarithms, the Prime Number Theorem is a good example why $e$ is the ``correct'' base of logarithms.
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Does your ten year old understand the slope of a line? If so, try explain to him what a tangent line is by some drawings and examples, and then tell him that the function $e^x$ has the property that the slope of its tangent line at $x$ is $e^x$. I mean, I know this is not a very creative answer, but it does indicate clearly what is so special about $e$.
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What about the basic (1+1/10)^10, (1+1/100)^100, and so on ?
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5$\begingroup$ My ten-year-old self would ask "so what?" and "why bother?" $\endgroup$ Commented Oct 9, 2010 at 12:06
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19$\begingroup$ My 18-year old calculus students also ask "so what?" and "why bother?". $\endgroup$ Commented Oct 9, 2010 at 19:56
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Thanks for all the answers. Some feedback...
Martin, I think the compound interest problem might be a bit beyond him. And J.M., I really don't know how he came across e, but I could ask him next time and report back. Qiaochu and others, I also think the secretary problem is a little bit hard.
Todd, I think your method is the best. e simply isn't as "obvious" as Pi, whichever way we choose, but linking it in to spirals is by far the most engaging. I'll give it a go and let you know how I get all.
Thanks all of you for your answers.
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1$\begingroup$ I don't know if my suggestion would actually "work"; I suspect it would work only if the child rather enjoys mathematics. But it could be fun to talk about logarithmic spirals (they are beautiful and occur in nature). I also remember some nice related visuals from Donald Duck in MathMagic Land. $\endgroup$ Commented Oct 9, 2010 at 14:13
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Here is how I began my explanation of $e$ to a class of (mostly) freshmen:
http://wnk.hamline.edu/~mjhardy/1170/homework/2nd.pdf
After #1 and #2, see the words "important punch-line".
#2 and #3 also have some material relevant to $e$.
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$\begingroup$ When you write that going back and forth between instantaneous rate of change and doubling time requires understanding e, wouldn't it be more accurate to say that it requires the knowledge of logarithms? $\endgroup$ Commented Oct 10, 2010 at 1:48
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$\begingroup$ It requires base-$e$ logarithms. For example, if the doubling time is 1 year, then the instantaneous rate of change equals the current size times $log_e 2$ per year. $\endgroup$ Commented Oct 10, 2010 at 2:06
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4$\begingroup$ Sorry, typo. I meant $\log_e 2$, not $log_e 2$. I regret the distress this must have caused...... $\endgroup$ Commented Oct 10, 2010 at 2:07
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I know a much simpler way; you just need to describe for him/her how to exponentiate.
Let $a > b \ge e$ be two numbers (for simplicity, let them be positive integers). The question is, which one is greater: $a^b$, or $b^a$? For instance, let $a=1000$ and $b=999$. Which one is correct: $1000^{999} > 999^{1000}$, or $1000^{999} < 999^{1000}$? (I assume that equality does NOT hold.)
It can be shown that $a^b < b^a$. In fact, for every $x \ge 0$, we have $e^x > x^e$ (assuming $x \ne e$).
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$\begingroup$ Sadeq, your penultimate statement is wrong: consider $a = 5$ and $b = 1$. In your final sentence, you either need to replace $>$ by $\geq$ or specify that $x \neq e$. $\endgroup$ Commented Oct 9, 2010 at 12:27
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$\begingroup$ @Todd: I think you didn't notice the condition $a,b>1$ at the beginning of the 2nd paragraph. Yet, your second point is correct: I added the assumption $x \ne e$. $\endgroup$ Commented Oct 9, 2010 at 16:10
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$\begingroup$ You're right, Sadeq, that I didn't notice that condition. However, the statement is still incorrect if $a, b > 1$ are arbitrary numbers (take $a = 5$ and $b = 1.001$). Or, better yet, take $a = 4$, $b = 2$. $\endgroup$ Commented Oct 9, 2010 at 17:57