# Defining Euler's number via elementary euclidean geometry (and a dimension limit)

Let $B_n$ be a closed ball in euclidean space $\mathbb{R}^n$, and consider the largest cube $Q_n$ contained in $B_n$. Then, let $C_n$ be a cube of maximal size that is contained in $B_n$ and disjoint from the interior of $Q_n$. Set $$r_n:=\frac{\text{vol }Q_n}{\text{vol }C_n}.$$ This ratio is independent of the choices of $B_n$ and $C_n$. It depends only on the dimension $n$. Now, if I did not completely screw up my calculations, it holds $$\lim_{n\to \infty} r_n=e^4,$$ where $e$ is Euler's number $2,718...$. I find this remarkable as it allows us, in principle, to define $e$ using only certain natural intrinsic geometric ratios. However, in contrast to the definition of $\pi$ as the intrinsic ratio between circumference and diameter of a circle, we need to pass to the dimension limit here, which arguably makes the approach to $e$ above much less elementary from a conceptual point of view.

In view of the situation above, I would like to know:

Which other elementary geometric approaches to Euler's number are there?

Is there a textbook or an essay that deals with different elementary geometric approaches to $e$ and includes the one above?

Edit: Maybe my question is too vague, but I hope that the spirit in which I asked it comes through. In view of the first answer I got, I would like to emphasize that I am aware of the "usual" definitions of $e$ that are found in standard textbooks and the fact that there are many definite or indefinite integrals of basic functions that could be used to define Euler's number. They represent an area under some curve, which allows one to view them as "geometric" objects. However, this is not what I mean, even though of course all volume and length measurements could be written as some integral, and all statements about intrinsic ratios could be formulated as statements involving some quantity being $1$. Thus, by elementary euclidean geometry, I mean the geometry of lines, circles, balls, cubes, cones, polygons, hyperbolas, parabolas, etc., not in a particular coordinate system but rather in an invariant fashion.

Edit 2: As pointed out by Dan Romik, there are related questions on stackexchange. However, all related questions at stackexchange I can find are of the open form "is there some geometric approach to $e$", and the best answers are very similar to Dan's. Yet I feel that the example above is conceptually somewhat different and I am concretely interested in approaches of a similar form, in contrast to asking a completely open question. Therefore, I would say that this question is not a duplicate of the related questions at stackexchange:

https://math.stackexchange.com/questions/382833/does-e-have-a-geometric-representation

https://math.stackexchange.com/questions/159707/is-there-any-geometric-way-to-characterize-e

• More simply, the volume of the unit ball in $n$-space is asymptotic to $${1\over\sqrt{n\pi}}\left({2\pi e\over n}\right)^{n/2}$$ – Gerry Myerson Oct 15 '15 at 21:57
• Ah, history repeats itself. Possible duplicate on Math.SE: "Is there any geometric way to characterize $e$?". But I dare say my figure is nicer than the one in that question... – Dan Romik Oct 16 '15 at 7:45
• The difference between this question and the related ones at Math.SE is that their authors were more or less satisfied with an answer equivalent to yours, but I am not ;-) – B K Oct 16 '15 at 9:52
• This question is a lot more interesting than I realized at first. It reminded me of an insight I had many years ago but had completely forgotten about, which is that $e$ is a lot less versatile of a mathematical constant than $\pi$. That is, $e$ doesn't actually appear naturally in mathematics in many different ways. When it does come up (e.g. the Poisson distribution in probability theory, and the various formulas one finds on Wikipedia and in Finch's book Mathematical Constant) this can almost always be traced back to two or three fundamental properties: ... – Dan Romik Oct 17 '15 at 6:34
• ... the infinite series definition, the limit $\lim_{n\to\infty} (1+1/n)^n$, and Stirling's formula. The one place I can think of where $e$ appears in a truly exotic and nontrivial way is in the number theoretic formula $e=\lim_{n\to\infty} (\textrm{lcm}(1,2,\ldots,n))^{1/n}$ that is one of the equivalent forms of the prime number theorem. So, I completely agree with @BK's motivation for the question: it would be extremely interesting to see some less familiar characterizations of $e$, if there even are any. – Dan Romik Oct 17 '15 at 6:34

$e$ is the unique number $>1$ such that the area in the plane bounded between the three lines $y=0$, $x=1$, $x=e$ and the hyperbola $y=1/x$ is equal to $1$.

• Of course I was aware of this characterization of $e$ and a hyperbola is quite an elementary object in geometry, but I feel that putting the hyperbola into a coordinate system and choosing $x=1$ as the lower bound of volume measurement reduces the "naturality" in this approach quite a bit. – B K Oct 15 '15 at 21:22
• @BK I agree. Nonetheless, you asked for an elementary geometric definition of $e$, so I provided one. :-) – Dan Romik Oct 15 '15 at 21:23
• No need for explicit coordinates. Let $\alpha$ and $\omega$ (as in $\alpha$bscissa and $\omega$rdinate) be the asymptotes of our hyperbola $H$, meeting at $O$. Thus $H$ consists of all points that, together with their projections to $\alpha$ and $\omega$ and with $O$, span a rectangle of unit area. Suppose lines $\ell$, $\ell'$ are parallel to $\omega$ and meet $\alpha$ at $P,P'$. Then $H,\ell,\alpha,\ell'$ enclose a unit area iff the ratio between $OP'$ and $OP$ is $e$. – Noam D. Elkies Oct 15 '15 at 22:37
• If ye suppose that $B_n$ contains the cube $Q_n$, then the edge of $Q_n$ would be $\sqrt{2/n}$. Taken to the $n$th power, this would go very tiny very fast. – wendy.krieger Dec 25 '16 at 13:58

Four ants are on the corners of a square, each facing its neighbor in the counterclockwise direction. At the same time each begins marching towards its neighbor, all at the same speed. After moving through one radian of angular measure about the center of the square, they are each closer to the center of the square than they were at the beginning by a factor of $e$.

Analytically, the ants trace out logarithmic spirals, describable as curves $t \mapsto \exp((-1+i)t) z_0$ where $z_0$ is the starting position.

• Surely, logarithmic spirals are very natural objects that even occur approximately in nature. Therefore, I like this answer very much. Yet I don't know whether one should consider logarithmic spirals as belonging to "elementary euclidean geometry", and even if one does, your example involves a dynamical process, so that euclidean geometry actually determines the problem only indirectly: it is only needed to formulate the rule how the ants should move. – B K Oct 16 '15 at 10:03
• @BK You could also view these trajectories (up to one radian) as limits of unions of line segments where the n-th union is formed by joining $z_k$ to $z_{k+1} = (1 + (-1 + i)/n)z_k$, starting at $z_0$ and ending at $z_n$. So if you are admitting limiting procedures as in your answer, then I'd think this would be admissible. – Todd Trimble Oct 16 '15 at 10:16

Carving Euler’s number in a carrot.

As noted by others, $e = 2.718...$ is transcedental, so we can’t construct this length with 'compass and straightedge'. So let's try it with 'carrot and knife'!

I will do this by carving the exponential function $f(\theta) = e^\theta$ on the mantle of a cylinder (a carrot).

1) Place the carrot under the $y-$axis, and position a knife on top of the line $y = x + 1$

2) Fixate the start of the blade at the intercept with the $x-$axis, such that this serves as a rotation point. With the start orientation of $45^{\circ}$, the knife initially cuts the mantle at $(x,y) = (0,+1)$:

3) Now start rotating the carrot in the counter clockwise direction. This makes that the knife will create a continuous carve on the carrot's mantle. As a result, the intersection point between knife and cylinder, will move away over the $y-$axis.

Claim: The resulting carve forms the exponential function $f(\theta) = e^{\theta}$.

Explanation

Note that the knife always cuts the carrot at the $y-$axis. This makes that for the tangent of the knife we have:

$$\frac{\Delta y}{\Delta x} = \frac{y-0}{0 - (-1)} = y$$

This yields a differential equation in the cylindrical coordinates of the carrot:

$$\frac{dy}{d \theta} = y(\theta)$$

Combined with the initial condition $y(0) = 1$, the solution of the differential equation $y’= y$ is indeed $y(\theta) = e^\theta$

The finish

Our cut $y=e^\theta$ intersects the line $\theta = 1$ at $(\theta, y) = (1, e)$. Connecting this point with $(\theta, y) = (1,0)$ creates a vertical segment with length $e = 2.71828...$

$\blacksquare$

PS: Rotating the carrot $180^{\circ}$, we get a length of $=e^{\pi}$ (Gelfond’s constant).

• This is actually quite similar to my answer. But nice illustrations! – Todd Trimble Dec 25 '16 at 12:10
• Just a question, did you already have pictures of this 3d model--as in you've explained this before--or did you do these just for this MO question? If it is the latter, I am impressed. – user78249 Dec 26 '16 at 1:25
• @james.nixon. I made these for an tanswer on Quora. The base was made in Matlab and some annotation in Inkscape. – Job Bouwman Dec 27 '16 at 18:39

As far as I know, $e$ was encountered the first time as the answer to the question, what would happen, if the intervals, in which interest is paid and reinvested, tend to zero or, for short: $$e := \lim_{n\to+\infty}\left(1+\frac{1}{n}\right)^n$$
The geometric interpretation is straight forward:

$e$ is the limit-volume of an $n$-dimensional cube with sides of length $1+\frac{1}{n}$

As hyper cubes of the desired side-length can be constructed with straight-edge and compass, and because hyper cubes are of elementary geometry, that should answer the question

• This is probably the simplest possible elementary geometric interpretation of $e$, and therefore appealing in its own way. However, it feels artificial and non-intrinsic to me that one has to consider exactly the side length $1 + \frac{1}{n}$ in the $n$-th dimension. Geometrically, there seems to be no natural reason to choose precisely these lengths, unless of course you already know that you want to define $e$ ;-) In contrast, the example in my question does not require any quantitative choices. – B K Oct 18 '15 at 18:53
• @BK hmmm, so the question would be, whether cubes with such sidelengths are the solution to some natural mathematical problem; which from a philosiphical point of view should be the case, because their volume equals an "universal" constant ;-) – Manfred Weis Oct 19 '15 at 3:06

Speaking of hypercubes (drawing on the comment by Manfred Weis), there is another perspective that is geometric--but also combinatorial.

Here is Euler's Number expressed a series of hypercubes each divided by n!, where n is the dimension of the hypercube.

$$e^{x} = \sum_{n=0}^{\infty} \frac{x^{k}}{k!} = \frac{x^{0}}{0!} + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \cdots$$

You'll notice a similarity to elementary problems in combinatorics: from an arbitrary set {a,b,c,d,e} where each element is in [0,x], how many ways can you form coordinates in $$\mathbb{R}$$3? It's just a combination with repetition problem. There are

$$\binom{n + 1 - r}{r}$$

such ways. Similarly, each term can be thought of the set of combinations of coordinates with repetition in each hypercube space.

• This is identical to the answer you posted earlier and then deleted, isn't it? Why would you do that? – Gerry Myerson Jun 4 '19 at 22:30
• I modified/edited the post to the extent where anyone reading the original comments would be confused by them. Original comments: 1) "There are 𝑛! ways to arrange the coördinates when all coördinates are distinct. – LSpice 2) I assumed you were referring to, say, changing (0,1,2) to (2,0,1) as "arranging the coördinates". In this case, even though the vectors, say, (0,1,0) and (0,0,1) are linearly independent, switching the y- and z-coördinates of (0,1,1)gives (0,1,1), which is still the same vector. (Indeed, as you mention, it has not 3! but 3!/(2!1!) distinct re-arrangements.) -LSpice – Atton Rand Jun 4 '19 at 23:21