Why is $e$ the number that it is? [duplicate]

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Why is $e$ the number that it is? Most of the irrational number that we learn about in school have something to do with geometry, like $\pi$ is the ratio of a circle's diameter to its circumference. Those numbers can also be derived using maths ($\pi = \lim_{n\to\infty}4\sum_{k = 1}^{n}\frac{(-1)^{k+1}}{2k - 1}$). On the other hand, $e$ is derived only from math, and finds no other immediate geometric basis. Why then, is $e$ the number that it is ($2.7182\ldots$) and not some other number?

EDIT: My question is more along the lines of why $e$ is $2.7182\ldots$ On an alien world where there is a different system of numbers, could there be a similar constant that has all the properties of $e$, but is not $2.7182\ldots$?

marked as duplicate by quid♦, Chill2Macht, user99914, Strants, Alex M.Jul 31 '16 at 21:01

• In many ways, $e$ has a relationship to the unit hyperbola that is similar to the relationship $\pi$ has to the unit circle. – Brian Tung Dec 2 '15 at 18:58
• @KimPeekII: The summation to infinity typically is defined as the limit of the partial sums, is it not? – Brian Tung Dec 2 '15 at 20:16
• Perhaps, historically, it may have first arisen in connection to the area under a hyperbola. But I'm not sure. – Akiva Weinberger Dec 2 '15 at 21:34
• @AkivaWeinberger: The development of the natural logarithm (which of course has $e$ has its base) had its origin, I seem to recall, in the quadrature of the hyperbola. – Brian Tung Dec 2 '15 at 22:02
• Related question: math.stackexchange.com/questions/26037/… – Gerry Myerson Jul 31 '16 at 9:22

Let's begin with a geometric interpretation that is, admittedly, a little contrived, but may give some satisfaction. Consider a circle, and construct a rectangle whose width is one of the circle's radii. If we extend the black lines outward until the two orange areas together have the same area as the purple semicircle, the length of the rectangle is $\pi$ times its width: Next, consider a right hyperbola (that is, one in which the asymptotes are at right angles). If we extend the black lines outward until each of the orange areas individually have the same area as the purple square, those black lines are each $e$ times as far out from the center as the side of the square: Now, some other observations:

Remember that hyperbolic sine and cosine are defined as

$$\cosh\phi = \frac{e^\phi+e^{-\phi}}{2}$$ $$\sinh\phi = \frac{e^\phi-e^{-\phi}}{2}$$

so that $e$ is as important to these functions as $\pi$ is to ordinary sine and cosine.

The unit circle $x^2+y^2 = 1$ can be parametrized as $(\cos\theta, \sin\theta)$, and then the area of the sector subtended at the center by the circle, from $0$ to $\theta$, is $\theta/2$. Similarly, the unit hyperbola $x^2-y^2 = 1$ can be parametrized as $(\cosh\phi, \sinh\phi)$, and then the area of the sector subtended at the center by the hyperbola, from $0$ to $\phi$, is $\phi/2$.

Equivalently, just as we have $\cos^2\theta+\sin^2\theta = 1$, we have $\cosh^2\phi-\sinh^2\phi = 1$.

The figure produced by a hanging chain of uniform density is called a catenary (from Latin catena "chain"), and is simply $y = \cosh x$: • An excellent answer! +1. This is a very beautiful construction that I've been trying to find – Yuriy S Mar 2 '16 at 17:17

The significance of $e$ is that it is the only number satisfying $\frac{d}{dx}e^x=e^x$.

$$e^x\ge x+1,\quad\text{for all }x$$ This uniquely defines $e$, and is my personal favorite property of $e$. (This is partly because it defines it without appealing to calculus.)

Basically, if you want to do inequalities with exponentials, you need $e$. (For example, what is the smallest value of $x^x$? The answer is $(\frac1e)^{1/e}$, at $x=\frac1e$.) A similar thing happens with trig functions and radians.

• I agree on the favorite property. I myself often try to answer e-related questions here solely from $\exp(x)\ge 1+x$ and $\exp(x+y)=\exp(x)\exp(y)$. – Hagen von Eitzen Dec 2 '15 at 21:38
• A while ago I figured out how to prove that:$$\ln2=1-\frac12+\frac13-\frac14+\dotsb$$just using those (and the squeeze theorem at the very end), without calculus. However, I haven't figured out how to do the same with $\ln2=\sum_{n=1}^\infty\frac1{n2^n}$. – Akiva Weinberger Dec 2 '15 at 21:49
• It is a bit illusory to say that this defines $e$… if you wish to prove the existence of a unique number $e$ that satisfies this property, calculus is all over the proof. But I like this nonetheless. – Lee Mosher Dec 3 '15 at 2:15
• @LeeMosher Well, not really. Pretty sure you just need Cauchy's intersection theorem, or maybe even just the squeeze theorem. A bit analysis-y, but not really "calculus all over it." – Akiva Weinberger Dec 3 '15 at 2:22

$e$ has a highly geometrical interpretation when you consider $e^{i \theta}$: it becomes a description of the idea of "rotation".

Additionally, $e$ occurs physically in a very wide range of situations: any time growth compounds with itself to cause further growth, there will be an exponential lurking beneath the surface. $e$ turns up in the real world all the time.

The definition of $e$ is as follows: $$\lim_\limits{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$$ and hence $$2<e<3$$

Apart from that, $e$ has other representations as well: $$e=\sum_{n=0}^\infty \frac{1}{n!}$$

And from the point of view of geometry, consider Euler's famous formula: $$e^{i\theta}=\cos \theta+i\sin \theta$$

This has immense importance in geometry of complex numbers, especially in case of "rotation".

If you consider the functions $a^x$, where $0<a<1$ or $a>1$, all these functions have the point $(0,1)$ on the graph (this is the only point common to all these graphs, so it is obviously an important point).

Only for $e^x$ does this function have slope exactly $1$ at $(0,1)$. The number $1$ is important as it is the so called multiplicative identity, $x\cdot 1=x$ for all $x$. So $e$ is important as being the specific base that makes $e^x$ the only function, among all $a^x$, that has slope $1$ at the point $(0,1)$. Slope $1$ is also geometrically significant as the only slope for which "rise"$=$"run".

I don't think there is one set reason as to "why" $e$ is what it is. $e$ is a number that satisfies many properties, much like $\pi$ and that is in a way $e$ is the number that it is. Along with the fine properties that the other answers have provided I'll include one of the first things Euler discovered about $e$ was the continued fraction: $e = 2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+ \cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{6 + \cdots}}}}}}}}$
Notice the pattern or even numbers. I find that to be extremely interesting about $e$.

• Douglas Hofstadter pointed out in his book Fluid Concepts and Creative Analogies that the initial $2$ doesn't fit the nice repeating $[\ldots, 1, 2n, 1, \ldots]$ pattern, but that blemish goes away if you reinterpret the continued fraction as $[1, 0, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, \ldots]$ instead. – Rahul Dec 3 '15 at 0:26

The identity (which can be seen as a definition) $e^{i\theta} = \cos\theta + i\sin \theta$ is geometrically significant. If you delve into the subject of complex analysis, you'll understand that $e^{i\theta}$ can be seen as a "rotation".

Not forget to mention:

• $$e^x = \sinh(x) + \cosh(x)$$
• $e$ is fundamental for the Theory of Representations in Group Theory, essential topic in Quantum field theory, quantum mechanics and particle physics
• $e$ arises also from the "compound interests" problem in economics
• It's fundamental for Fourier transforms, Stirling approximation and Gaussian curves (so that it's fundamental in Data Analysis and Probability and Statistics)

$e$ is like matrices. Or sets. When you first learn them, you think, "Err...so what?" As you learn more mathematics, you find yourself using these things more and more, until you can't imagine doing mathematics without them.

For $u>0$ and $f(u)=\int_1^u (1/x)dx$ , a geometric representation of $e$ is that $f(e)=1$. The region bounded by the $x-$axis, the line $y=1$,the graph of $y=f(x)$, and the line $y=u$, has area $1$ when $u=e$. This can be taken as a definition of $e$, analogous to the definition of $\pi$ as the area of a semi-circle of radius $1$.