There are so many available bases. Why is the strange number $e$ preferred over all else?

Of course one could integrate $\frac{1}x$ and see this. But is there more to the story?

Martin Sleziak
  • 50,316
  • 18
  • 169
  • 342

6 Answers6


Differentiation and integration is precisely why it is considered natural, but not just because $$\displaystyle\int \frac{1}{x} dx=\ln x$$

$e^x$ has the two following nice properties

$$ \frac{d}{dx} e^x=e^x $$

$$ \int e^x dx=e^x+c $$

If we looked at $a^x$ instead, we would get:

$$\frac {d} {dx} a^x= \frac{d}{dx} e^{x\ln(a)}=\ln(a) \cdot a^x$$

$$\int a^x dx= \int e^{x\ln(a)} dx=\frac{a^x}{\ln(a)}+c$$

So $e$ is vital to the integration and differentiation of exponentials.

draks ...
  • 17,825
  • 7
  • 59
  • 171
  • 8,703
  • 7
  • 50
  • 79

If you know some linear algebra, then here is an abstract reason: $e^x$ is the unique eigenvector of eigenvalue $1$ of the derivative $D$ acting on, say, the space of smooth functions on $\mathbb{R}$. Why is this important? The study of solutions of linear differential equations with constant coefficients is equivalent to the study of nullspaces of operators which are polynomials in $D$, e.g. operators of the form $\sum a_k D^k$. Any such operator automatically commutes with $D$, so this nullspace splits up into eigenspaces of $D$. That's why solutions to linear differential equations with constant coefficients can be expressed as sums of complex exponentials. The choice of $e$ makes it particularly easy to see what the eigenvalue is: the eigenvalue of the eigenvector $e^{\lambda x}$ is $\lambda$.

Qiaochu Yuan
  • 359,788
  • 42
  • 777
  • 1,145
  • I would think that very few function are eigenvectors of the Derivative operator. In fact, I'd be curious to know if there are any others besides the exponentials and even some sums of sinusoidals/hyperbolics. – crasic Nov 09 '10 at 08:36
  • 2
    @crasic: it's just the exponentials. However, functions of the form x^k e^{lambda x} are _generalized_ eigenvectors of the derivative operator, so these are the full set of functions you need to solve homogeneous constant-coefficient linear ODEs. – Qiaochu Yuan Nov 09 '10 at 09:07
  • 1
    This is a nice explanation of the relation between *D* and exponentials, but I don't see how it answers the original question, i.e. why is eigenvalue 1 more important than any other eigenvalue? Except for the fact that 1 is a nice number (being a multiplicative identity and all), I don't see any rational argument why *e* should be preferred over any other base. – Marek Nov 09 '10 at 13:26
  • 2
    @Marek: e^x is the trivial representation, and the trivial representation is always special (for example it is the identity object in the monoidal category of representations). – Qiaochu Yuan Nov 09 '10 at 15:21
  • @Qiaochu: You must be using some strange definitions, because in my book trivial representation is always the one that sends everything to identity. In this case it's e^{0x} = 1. Another way to see this is that characters of (R, +) e^{\lambda x} form a dual Potryagin group isomorphic to (R, +). – Marek Nov 09 '10 at 16:46
  • 2
    @Marek: I'm not talking about any action of R. I'm talking about certain subrepresentations of the representation of C[X] on C^{\infty}(R) where X acts as the derivative. The trivial representation is the one where X acts trivially (that is, is sent to the identity _operator_), and it occurs with multiplicity 1 and is spanned by e^x. – Qiaochu Yuan Nov 10 '10 at 00:42
  • @Qiaochu: Oh, I didn't know about this. Nice. – Marek Nov 10 '10 at 08:43
  • I would say that this is an extension of the naturalness of `e`, but by no means what makes it natural. Everyone can agree that `e` is very useful. Most answers here say the same thing (as is the intent of math), but @JustinL put it succinctly, "the rate of change of the exponential curve at any point is equal to the y value of the curve at that point." `e` is the essence of exponentiation; its rate of change is itself. – J.Money Jan 10 '14 at 02:31
  • This is probably a basic question. Why is it true that if two operators commute, then the nullspace of one splits into the eigenspaces of the other? Can someone point me to a reference? – Matthew Kvalheim May 05 '16 at 19:50
  • @Matthew: this is a straightforward computation. If $MN = NM$ and $Mv = 0$, then $MNv = NMv = 0$. Hence $N$ preserves the nullspace of $M$. – Qiaochu Yuan May 05 '16 at 20:19
  • Ah, simpler than I thought it would be. Thanks @QiaochuYuan – Matthew Kvalheim May 05 '16 at 20:31
  • This just unified like 3 different concepts into one for me. Nice. I didn't know how to express it, but there seemed to be something behind the derivative of e^x "following it's own line". – user Dec 12 '20 at 10:55

The wikipedia article on e tells a bit of the story.

One example is an account that starts with 1.00 and pays 100% interest per year. If the interest is credited once, at the end of the year, the value is 2.00; but if the interest is computed and added twice in the year, the 1 is multiplied by 1.5 twice, yielding 1.00×1.5² = $2.25. Compounding quarterly yields 1.00×1.254 = 2.4414…, and compounding monthly yields 1.00×(1.0833…)12 = 2.613035….

Bernoulli noticed that this sequence approaches a limit (the force of interest) for more and smaller compounding intervals. Compounding weekly yields 2.692597…, while compounding daily yields 2.714567…, just two cents more. Using n as the number of compounding intervals, with interest of 100%/n in each interval, the limit for large n is the number that came to be known as e; with continuous compounding, the account value will reach 2.7182818…. More generally, an account that starts at $1, and yields (1+R) dollars at simple interest, will yield eR dollars with continuous compounding.

Additionally, it is the base of the exponential function y = k^x, finding a specific value for k where d/dx k^x = k^x. That is, the rate of change of the exponential curve at any point is equal to the y value of the curve at that point.

Justin L.
  • 13,646
  • 22
  • 60
  • 73

The Wikipedia article about e lists many properties of the constant that make it naturally occurring.

I think the biggest reason it is natural when it comes to exponentiation/logarithms is that it is the only number that satisfies

$$ \frac{d}{dt} e^t =e^t $$ while every other number satisfies

$$ \frac{d}{dt} a^t = c \cdot a^t$$ where $c$ is some constant, different than 1. This makes it "normalized" in a sense.

Ellie Kesselman
  • 453
  • 7
  • 24
Eric O. Korman
  • 18,051
  • 3
  • 52
  • 82

I'm surprised I never answered this; maybe I was deterred by the fact that several other answers are here.

One short answer is this: An exponential function $y=a^x$ grows at a rate proportional to its present size, but only when the base is $e$ does it grow at a rate equal to its present size. In other words $$ \frac{d}{dx} a^x = \left(\text{constant}\cdot a^x \right), $$ but only when $a=e$ is the "constant" equal to $1$.

The number $a=2$ is too small for this to happen. To see that consider $$ \frac{d}{dx} 2^x = \lim_{h\to0} \frac{2^{x+h}-2^x}{h} = \lim_{h\to0}2^x\frac{2^h-1}{h} $$ This last limit is equal to $\displaystyle 2^x \lim_{h\to0}\frac{2^h-1}{h}$. That step can be done because $2^x$ is a "constant", but in this instance, "constant" means "not depending on $h$". Then observe that $\displaystyle\lim_{h\to0}\frac{2^h-1}{h}$ is a "constant", where "constant" now means "not depending on $x$".

So $$ \frac{d}{dx} 2^x = \left(\text{constant}\cdot 2^x\right). $$ But what number is this "constant"? Notice that as $x$ increases from $0$ to $1$, $2^x$ increases from $1$ to $2$, so the average slope on that interval is $\dfrac{2-1}{1-0}=1$. Since the curve gets steeper as $x$ increases, it's not yet that steep at $x=0$. Its slope at $x=0$ is $\left.\dfrac{d}{dx}2^x\right|_{x=0}=\left(\text{constant}\cdot2^0\right)$, so that "constant" must be less than $1$.

A similar argument shows that if $4$ is used as the base, the "constant" is more than $1$. This is done by using the interval from $-1/2$ to $0$ instead of the interval from $0$ to $1$.

So $2$ is too small, and $4$ is too big, to be the natural base. $e$ must be somewhere between $2$ and $4$. In a similar way one can show that $3$ is to big, but that's where the previously simple arithmetic gets messy. Use the interval from $-1/6$ to $0$ for that.

Similarly with logarithms: \begin{align} \frac d {dx} \log_6 x & = \frac{\text{constant}} x, \\[12pt] \text{ and } \quad \frac d {dx} \log_e x & = \frac{\text{constant}} x, \text{but this time, the constant is 1.} \end{align}

What is natural about $e$ is the same thing that is natural about radians: $$ \frac d {dx} (\text{sine of $x$ degrees}) = \Big((\text{cosine of $x$ degrees}) \times \text{constant} \Big) $$ but only when radians are used is the "constant" equal to $1$. $($With degrees the constant is $\pi/180.)$

Michael Hardy
  • 1
  • 30
  • 276
  • 565

If you consider all exponential equations $a^x$, they all have $y$-intercept $(0,1)$. If you wanted to specify an archetypal exponential equation to refer to as you work through Calculus, a natural choice would be to choose the one whose tangent line at $(0,1)$ has slope 1. The equation $e^x$ is the unique exponential equation with that property.

Austin Mohr
  • 24,648
  • 4
  • 62
  • 115