Why is $e$ so special?

Question

The number $e$ (and the exponentiation function $e^x$) appears in so many places in mathematics and engineering. There seem to be a multitude of applications of it. I want to know why.

Answer 1

$e$ is the positive real number which fulfils differential equation \begin{equation} \frac{d}{dt} e^t = e^t \end{equation} Also, in engineering problems, the number $e$ appears in the solutions of systems of differential equations (see e.g. the wikipedia article about Linear differential equations).

Answer 2

Why is e so special ?

Because very many of humanity's age-old mathematical interests ultimately converged towards it:

Arithmetic:

• Addition begets multiplication; multiplication begets exponentiation; exponentiation begets e, since by studying it we inevitably arrive at the conclusion that this number is its most natural base.

• $\sqrt[\large^n]{\text{LCM}(1,2,3,\ldots,n)}~$ tends to e as n tends towards infinity.

Geometry:

• Circles and hyperbolas have been studied since ancient times; e is to the latter what $\pi$ is to the former.

Finance:

• Examining the way in which banking interests are computed leads us to discovering the same quantity.

Calculus:

• The harmonic series has been studied since ancient times; its continuous equivalent is $\displaystyle\int\frac1x~dx$ $=\log_ex$.

• The solution to $f(x)=f'(x)$ is $a~e^x$, meaning that the exponential function is immune to the operations of differentiation and integration.

Answer 3

Exponentials pop up in linear differential equations, which govern innumerable physical phenomena. Such equations appear every time the variation of a physical quantity is proportional to that same physical quantity.

Let us take the discharge of a capacitor in a resistor: due to the resistor, the current is proportional to the tension, and due to the capacitor the tension is proportional to the load. Lastly, the capacitor empties at a speed proportional to the current.

This is summarized by the archetypal equation below:

$$\frac{du}{dt}=-u.$$

This equation is solved as

$$\frac{du}u=-t,$$ then $$\int\frac{du}u=-\int dt,$$ then $$\log_e(u)=-t$$ then $$u=e^{-t}.$$

It expresses that the capacitor will discharge exponentially, i.e. after $1$ unit of time, only $1/e=37\%$ of the initial load remains, after two units of time, $1/e^2=14\%$, after three units of time, $1/e^3=5\%$... The decrease is fast and follows a geometric progression.

This is a typical behavior of systems, that move from one state to another following a transient stage with an exponential decay. The differential equation shows you how the exponential function appears. It also shows you that $e$ is a natural base of exponentials and logarithms, as it is the only base such that

$$(\log_b(x))'=\frac1x$$and similary$$(b^x)'=b^x.$$

$e$ is a natural "unit" in the same sense that $2\pi$ is a natural angular unit for the trigonometric functions, as

$$(\sin(x))'=\cos(x)$$

is valid in radians, i.e. when a full turn is $2\pi$.

(By contrast, $(2^x)'=0.69314718\cdots 2^x$, and $(\sin_°(x))'=0.01745329\cdots\cos_°(x)$ in degrees.)

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Actually, all the elementary functions that you find on a calculator are closely related to the exponential: the exponential $e^x$ itself, its inverse $\ln(x)=\log_e(x)$, the exponential and logarithms in other bases, $b^x$ and $\log_b(x)$, the powers $x^y$, computed from $x^y=e^{y\ln(x)}$; then the trigonometric functions constructed from the imaginary exponential $e^{ix}=\cos(x)+i\sin(x)$, the functions derived from them $\tan,\sec,\csc,\cot$ and their inverses $\arccos,\arcsin,\arctan,\cdots$. Ditto for the hyperbolic functions.

As logarithms and antilogarithms can be used to perform multiplies and divides, a great deal of the maths can be done with $+, -, \ln(z),e^z$ only ! ($z$ to denote complex numbers.)

For example,

$$\arccos(x)=-i\ln\left(x+ie^{e^{\ln\left(\ln\left(1-e^{e^{ln(ln(x))+ln(2)}}\right)\right)-\ln(2)}}\right).$$

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(This section is informal.)

To compute the value of $e$, let us use the differential equation. As $$(e^x)'\Big|_{x=0}=\lim_{x\to0}\frac{e^x-1}x=e^0=1,$$ we have $$e^x\approx1+x$$ for small $x$.

Then $$e=(e^{1/n})^n\approx\left(1+\frac1n\right)^n,$$ and $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$

Also, by the binomial theorem,

$$\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}\cdots\approx1+1+\frac12+\frac1{3!}\cdots$$ and in the limit, $$e=\sum_{k=0}^\infty\frac1{k!}.$$

Answer 4

The idea of exponential growth or decay is natural: certain continuous quantities increase $10\%$ every year, other quantities get halved in the course of $10\,000$ years, etcetera. Given a unit of time there are still "slow" and "fast" exponential increases: a quantity can double in one unit of time, or it can increase by a factor of $100$ in one unit of time. Therefore we need a unit for the "speed" of exponential growth. We arrive at it in the following way: Among all exponential growth curves $t\mapsto f(t)$ with $f(0)=1$ there is exactly one which has slope $1$ at $(0,1)$. The base for this particular $f$ is a certain number $>1$. This number is called $e$, its value turns out to be about $2.718$. In this way the standard exponential function is $f(t)=e^t$.