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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.

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    @IanMateus Cê tem umas perguntas muito boas.2012-09-13

11 Answers 11

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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."

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    @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
20

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

  • 1
    This is a far cry from sqrt and $pi$2016-07-08
16

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/

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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
14

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.

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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
10

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 $\frac{PP_0}{QQ_0}=e$.

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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

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

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    @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 $e=\sum_{n\ge 0}1/n!$.

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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
    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