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$\begingroup$

Let me explain it better: after this question, I've been looking for a way to put famous constants in the real line in a geometrical way -- just for fun. Putting $\sqrt2$ is really easy: constructing a $45^\circ$-$90^\circ$-$45^\circ$ triangle with unitary sides will make me have an idea of what $\sqrt2$ is. Extending this to $\sqrt5$, $\sqrt{13}$, and other algebraic numbers is easy using Trigonometry; however, it turned difficult working with some transcendental constants. Constructing $\pi$ is easy using circumferences; but I couldn't figure out how I should work with $e$. Looking at enter image description here

made me realize that $e$ is the point $\omega$ such that $\displaystyle\int_1^{\omega}\frac{1}{x}dx = 1$. However, I don't have any other ideas. And I keep asking myself:

Is there any way to "see" $e$ geometrically? And more: is it true that one can build any real number geometrically? Any help will be appreciated. Thanks.

  • 12
    If by "geometrical" you mean compass and straightedge, then no it is not possible since $e$ is a transcendental number.2012-06-18
  • 11
    Yes, kind of it. But $\pi$ is a trancendental number, and it can be "seen" in a circumference of radius $\frac{1}{2}$.2012-06-18
  • 8
    Then I would ask you to define the constraints of when something is "seen" :). I could say that from your circle you can "see" $e$ because a circle is parametrized in the complex numbers as $e^{i \theta}$.2012-06-18
  • 2
    $e$ is the value of $w$ such that the region under the reciprocal curve, $y=1/x$, from $x=1$ to $x=w$, has area $1$. (That is, $\int_{1}^{w} \frac{1}{u} du = 1$.) Unlike your $\int \log$ example, this approach doesn't implicitly use $e$ to define the region's upper boundary.2012-06-18
  • 2
    @tomcuchta In my question, "seen" is when a constant appears in a geomatrical construction, using compass and straightedge -- as you said! I'm asking it because of the fact that there is a transcendental number $\left(\pi\right)$ which appears in this construction. I'm thinking in words to put it in a clear way, but I couldn't find them. It is kind of "intuition".2012-06-18
  • 0
    @DayLateDon Didn't note it, thanks!2012-06-18
  • 1
    @IanMateus Cê tem umas perguntas muito boas.2012-09-13

10 Answers 10

51

For a certain definition of "geometrically," the answer is that this is an open problem. You can construct $\pi$ geometrically in terms of the circumference of the unit circle. This is a certain integral of a "nice" function over a "nice" domain; formalizing this idea leads to the notion of a period in algebraic geometry. $\pi$, as well as any algebraic number, is a period.

It is an open problem whether $e$ is a period. According to Wikipedia, the answer is expected to be no.

In general, for a reasonable definition of "geometrically" you should only be able to construct computable numbers, of which there are countably many. Since the reals are uncountable, most real numbers cannot be constructed "geometrically."

  • 7
    I read your last paragraph as implying that $e$ is not, or may not be computable. I am sure you did not mean to imply this.2012-06-18
  • 6
    @Mark: I do not see where I make that implication. Claiming that all geometrically constructible numbers are computable is not claiming that all non-geometrically constructible numbers are not computable.2012-06-18
  • 3
    It's because you didn't indicate that you were no longer talking about $e$.2012-06-18
  • 9
    @Mark: I think it was quite clear from the words "in general" that Qiaochu was answering the last question of the OP and not talking about $e$ in particular.2012-06-18
19

The area beneath the reciprocal function $$\frac{1}{x}$$ from $x=1$ to $x=e$ is $1$. Though this isn't really geometric like you want, it is still a clear way to see $e$ physically.

e

  • 6
    I would call this a way of *recognizing* $e$, not *constructing* $e$.2012-06-18
  • 5
    I'd say it gives us a way to construct an approximation to $e$ much like inscribing/circumscribing polygons in a circle allows us to construct an approximation to $\pi$.2012-06-18
  • 1
    This is a far cry from sqrt and $pi$2016-07-08
15

Debeaune asked Descartes this problem in a letter in 1638:

Consider a curve $y=f(x)$. Lets consider the tangent line $t(x)$ through the point $(x_{0},y_{0})$ which would look like $t(x) = y_{0} + f'(x_{0})\cdot(x-x_{0})$. What curve has the property that, every such tangent line intersects the $x$ axis at $x_{0}-1$, i.e., $$ t(x_{0}-1)=0 $$ What curve can do this? Only $y=C\exp(x)$...where $C$ is some nonzero constant.

For a thorough derivation, see http://pqnelson.wordpress.com/2012/06/03/exponential-function/

  • 0
    I would call this a way of *recognizing* $e$, not *constructing* $e$.2012-06-18
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    What's the distinction between the two? Using calculus to solve this problem explicitly constructs the curve $y=\exp(x)$, and this construction is **unique** when we specify $(0,1)$ lies on the curve.2012-06-18
  • 1
    How would you use this (nice) mathematical fact to construct $e$ (say a line segment of length $e$) from scratch? You can use any computer you want, as long as it doesn't already know what $e$ is. Does your method improve on constructing infinitely many line segments of lengths $1/n!$ and laying them end to end?2012-06-18
  • 0
    But isn't that going backwards, starting with one definition of $e$ then trying to concoct geometric problem? Debeaune's problem naturally gives us the limit definition for $\exp(x)$, from a given geometric problem, upon using basic differential calculus...I still fail to see how this is "recognizing" $e$ instead of "constructing" $e$...2012-06-19
  • 1
    @GregMartin: Such a construction of $e$ is easy. Have your computer graph _any_ exponential function, say $y=f(x)=2^x$. Now pick some $x_0$, draw the tangent through $(x_0,f(x_0))$ on the graph, intersect with $y=0$ to find $x_1$; now $f(x_0)/f(x_1)=e$, where $f(x_0)$ and $f(x_1)$ are lengths of segments easily found in the figure.2012-10-11
  • 1
    @Marc: this is a great idea ... but how does one construct the curve $y=2^x$, without using the exponential function? Again, I don't consider a limiting process a construction (else we can just construct $e = \sum 1/n!$ directly).2012-10-18
  • 0
    @GregMartin: Since the ratio $e:1$ is not constructible by straightedge-and-compass, you must of course do something else (I'm not trying to do the impossible). I use _an_ exponential curve, which is not conceptually worse than using a hyperbola: you can only actually construct finitely many points of it in finite time (but there is a dense subset of points that you _can_ construct). You cannot actually construct a tangent to the exponential curve, gain that would contradict that $1:e$ is non-constructible (here the hyperbola is different). My point is that you don't need the function $\exp$.2012-10-18
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    ... And just to make clear what I meant above: for every rational $\alpha$ you can construct the point $(\alpha,2^\alpha)$ by straightedge and compass.2012-10-18
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    Right; my point is that since you can only construct a dense set of points on the curve $y=2^x$, then you can only construct better and better approximations to $e$, not $e$ itself. And if we were going to settle for approximations, why not just use truncations to $\sum 1/n!$?2012-10-20
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    Also, I think it's true that any point that can be "constructed" using a straightedge, compass, and hyperbola can also be constructed using just a straightedge and compass. Assuming that's true, there is a real conceptual difference between hyperbolas and exponentials.2012-10-20
13

You can't build any real number geometrically. They aren't even all computable. If you don't want to consider functions, you could (this is kind of cheating) look at $\lim_{n\rightarrow\infty} (1+\frac 1 n)^n$ as the volume of a suitably sized hypercube as the dimension increases.

  • 0
    Sorry, but I would ask you why one can't build every real number geometrically.2012-06-18
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    It can get somewhat hairy, but basically any geometric construction is a finite algorithm (series of steps) of some kind. And by http://en.wikipedia.org/wiki/Computable_number most real numbers are not computable, meaning that they cannot be constructed with any such algorithm, geometric or not.2012-06-18
  • 0
    Going to take look at it. Thanks.2012-06-18
  • 0
    It's a tricky argument. Paying careful attention to how one means the words, I believe it's actually consistent for there to be a construction of every real number, despite there not being an (internal) bijection between (internal) constructions and real numbers within a set theoretic universe. Look at Skolem's paradox to see how such arguments can go horribly awry.2012-06-18
  • 0
    So, you can't build $\pi$ geometrically? Or $1$? Or $\sqrt{2}$? Your first sentense is exactly the same as 'you can build *no* real number geometrically'. Please, be more careful with your wording2016-03-02
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    @YuriyS he is probably negating "you can build any" rather than negating "you can build some" but this is not your typical use of "any" imho.2016-12-22
9

I'll take my comment at the answer of Alex Nelson and make it into a separate answer, although it is really just the same answer reformulated. But I'll start pre-empting the predictable comment "I would call this a way of recognizing $e$, not constructing $e$" by countering that, even if one proclaims that constructing a point at a given distance along a curved line is a valid operation, one still cannot construct $\pi$ with ruler and compass either: one can only recognise it as the distance around a circle of diameter $1$ needed to get to the point diametrically opposite to the starting point.

Obviously we need some non-ruler-and-compass ingredient to construct $e$. I'll take this to be the graph of some exponential function, together with its unique asymptote: say in some coordinate system in which that asymptote is the $x$-axis we are given the set of points $(x,a^x)$ for some $a>1$ (neither the unit length of the coordinate system nor the value of $a$ need to be known; the $x$-axis is of course determined by the graph, but I'd have difficulty giving a construction of it).

Here is the construction: pick a point $P$ on the graph, find the point of intersection $Q_0$ of the tangent line to the graph at $P$ with the $x$-axis. Then taking perpendiculars to the $x$-axis through $P$ and $Q_0$ which intersect the $x$-axis in $P_0$ respectively the graph in $Q$, one has $\frac{PP_0}{QQ_0}=e$.

  • 0
    I captured most of what you said but I don't know why the last procedure you have described in the last paragraph results in (e)? Thanks.2012-10-11
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    @EmmadKareem: There are two way to see this. The lazy way is to argue that up to horizontal scaling and translation, all graphs of exponential function are the same, so it suffices to check for the graph of $x\mapsto e^x$. The assiduous way is to verify that the tangent at $(x_0,a^{x_0})$ has slope $a^{x_0}\ln a$, therefore passes through $(x_1,0)$ where $x_1=x_0-\frac1{\ln a}$, so that $a^{x_1}=a^{x_0}a^{-1/\ln a}=a^{x_0}e^{-1}$2012-10-11
5

Another approach might be finding a polar curve such that it's tangent line forms a constant angle with the segment from $(0,0)$ to $(\theta,\rho(\theta))$. The solution is the logarithmic spiral, defined by

$$\rho =c_0 e^{a\theta}$$

  • 0
    I would call this a way of *recognizing* $e$, not *constructing* $e$.2012-06-18
  • 0
    @GregMartin This is definitely a construction and not just recognizing $e$. If you really want the value $e = 2.71828\ldots$, then just take $c_0 = 1$ and $a\theta=1$. Ensuring a constant angle for the tangent is a concrete procedure, just like involutes, evolutes, roulette, etc. Compass-and-straightedge is an idealization (of a physical process) that can never be achieved and only approximated; so are these. Confining "geometric construct" to merely compass-and-straightedge is an unnecessarily narrow view.2017-12-02
4

Plot the curves $y = a^x$ for $a > 0$. As $a$ gets larger, you find the slope of the tangent line is larger. If $a < 1$, this slope is negative. There is exactly one value for which the slope is 1, and that is $e$.

You can use this to define $e$ and derive the fact that $e=\sum_{n\ge 0}1/n!$.

  • 0
    I would call this a way of *recognizing* $e$, not *constructing* $e$.2012-06-18
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    Do you mean the slope at $x=0$?2012-10-11
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    Yes, Henry, at zero.2012-10-11
-1

Here is a quote from the Wikipedia article on necklaces in combinatorics. It suggests a way to visualize the inverse of e.

Necklace example[edit] If there are n beads, all distinct, on a necklace joined at the ends, then the number of distinct orderings on the necklace, after allowing for rotations, is n!/n, for n > 0. This may also be expressed as (n − 1)!. This number is less than the general case, which lacks the requirement that each bead must be distinct. An intuitive justification for this can be given. If there is a line of n distinct objects ("beads"), the number of combinations would be n!. If the ends are joined together, the number of combinations are divided by n, as it is possible to rotate the string of n beads into n positions.

-1

From the hyperbolic identity:

          cosh(x) + sinh(x) = e^x  ...in general, 

we have

          cosh(2) + sinh(2) = e^2  ...in particular. 

If we take individually, the square of the square root of each of the first two terms of the second equation, we have a Pythagorean identity, where the sides are:

F = √(cosh(2)), G = √(sinh(2)), and the hypotenuse of the right-angled triangle, H = e.

I hear you say, not as good as π, the ratio of the circumference of a circle to its diameter, since one can measure the circumference and the diameter, by some means. One needs the formula or Taylor series for cosh(2) and sinh(2).

(My original edit here was pointless. It involved the additional identity:

         1  +  sinh^2(x)  =  cosh^2(x) 

which is also a Pythagorean identity, but since it can be derived from the previous case, and vice versa, it was pointless in calculating a value for 'e').

N.B. 'e', the base of natural logarithms, was created as a valid idea in calculus, to make differentiation, and therefore integration, easier. It is not a geometrical constant, but a mathematical constant, and is the limit of:

                      (1 + t)^(1/t), as t tends to 0 

which occurs in the differentiation of log(x).

-1

EDIT: To my answer (under the line) I am gonna add pictures I have draw- I construct an aprroximation of $e$ into 5th iteration on the Number Line, where:

$$L_0=a_0=1 $$ $$L_{n+1}=L_n+a_{n+1} ; a_{n+1}=\frac{a_n}{n+1} $$

enter image description here Altenative we can write:

$$L_2=2.5 ;a_2=0.5 $$ $$L_{n+1}=L_n+a_{n+1} ; a_{n+1}=\frac{a_n}{n+1} $$

Here I tried to approximate between $[2;3]$ (into 6th step): enter image description here enter image description here

Old Answer:

---Introduction---

A) We gonna use the Taylor series for $e$ and multiply by $A$:

$e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...=1+\frac{1}{1}+\frac{1}{1*2}+\frac{1}{1*2*3}+\frac{1}{1*2*3*4}+...$

$Ae=A(1+\frac{1}{1}+\frac{1}{1*2}+\frac{1}{1*2*3}+\frac{1}{1*2*3*4}+...)$

In special case: $A=1$ (leght unit)

We can lead $Ae$ to this form:

$Ae=A(1+\frac{1}{1}(1+\frac{1}{2}(1+\frac{1}{3}(1+\frac{1}{4}(1+...)...)...)$ (Formula 1)

B) Optional: How a line segment can be devide via ruler and compass: Intercept theorem

C) (Formula 1) will be explained as an Algorithm plus picture .

---The Algorithm---

Short :

Start:

  • Have a line segment $a_0$ with a length of $A$ .

Loops:

I) Duplicate $a_{n-1}$ and divide the new segment into $n$ equal parts. Pick one element as $a_n$.

II) Add $a_{n-1}$ to the older line segments: $L_{n-1}=a_{n-1}+\displaystyle\sum_{i=0}^{n-2}a_i$

Long :

  1. Have a line segment $a_0$ with a length of $A$ .

1.1. Duplicate $a_0$ and divide the new segment into $1$ equal parts .

1.2. Pick one element as $a_1$.

1.3. Add $a_0$ to the older line segments: $L_0=a_0$

2.1. Duplicate $a_1$ and divide the new segment into $2$ equal parts .

2.2. Pick one element as $a_2$.

2.3. Add $a_0$ to the older line segments: $L_1=a_1+a_0$

3.1. Duplicate $a_2$ and divide the new segment into $3$ equal parts .

3.2. Pick one element as $a_3$.

3.3. Add $a_2$ to the older line segments: $L_2=a_2+a_1+a_0$

.

.

.

n.1. Duplicate $a_{n-1}$ and divide the new segment into $n$ equal parts .

n.2. Pick one element as $a_n$.

n.3. Add $a_{n-1}$ to the older line segments: $L_{n-1}=a_{n-1}+(a_{n-2}+...+a_2+a_1+a_0)$

---Picture--- enter image description here

  • 0
    Wish the downvoter had left a comment why.2018-09-09
  • 0
    I didn't downvote but perhaps it was because of the use of many colors, specially the use of light green in your handmade calculations. You should check this `https://math.meta.stackexchange.com/questions/4195/on-the-use-of-color-in-equations` it basically mention the complications for colorblind people, when using colors.2018-10-01