Why is $e$ so ubiquitous?

Question

$e$ is one of the most important numbers in our universe, it is everywhere. When I try to find out why the most common explanation is some reverse-engineered physics or finance problem. But these are just one off examples of why $e$ is important, they fall short of illuminating the origin of its significance. What I'm looking for is some fundamental definition of $e$ that explains it's significance and omnipresence, something akin to $\pi$ relating circumference and diameter.

Thanks, I hope I'm being clear.

EDIT*: A common saying is "$e$ is the most natural base." What does that mean?

I'm not much of a mathematician, but I think what @Tac-Tics said is another way of saying that if you are looking for some function $f(x)$ that is equal to its own derivative with respect to $x$, the answer is $f(x)=e^x$. — Solomon Slow, Aug 09 '16 at 17:14
$e$ is not important. What's important is the function $e^x$, and briefly it's important because it's its own derivative. I wrote a Quora answer about this that you can try searching for; not sure where it is at the moment. — Qiaochu Yuan, Aug 09 '16 at 17:15
Look here, as well as at the duplicate it points to. The answers collectively provide some insight into the ubiquity of $e$. — Brian Tung, Aug 09 '16 at 17:17
@Qiaochu Yuan: I would go so far as to say that $e^x$ (and its generalizations) is the most important function in mathematics. Besides the things people are saying here, it encompasses trigonometry (via the Euler identity), and it's important in Lie groups (see here), and it shows up in solving a linear system of ODEs, and . . . — Dave L. Renfro, Aug 09 '16 at 19:01

score 7 · Answer 1 · answered Aug 09 '16 at 17:19

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$e^{\lambda x}$ is the solution to the equation $Df=\lambda f$. This is equivalent to saying that $e^{\lambda x}$ is an eigenfunction of the differentiation operator D for any value $\lambda \in \mathbb{R}$. Now how ubiquitous is differentiation to physics?

answered Aug 09 '16 at 17:19

Wavelet

1,299

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That's really not very helpful, dude. – BoddTaxter Aug 09 '16 at 17:22
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Why not? $e$ is just a number. It doesn't have any magic qualities that you seem to be looking for, at least aside from this relationship. – Wavelet Aug 09 '16 at 17:26
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Because it is just a bunch of jargon. I honestly have no idea what any of that means, and I have a feeling I wouldn't need to ask this question if I did so I don't know who your answer is even meant for. "$e$ is just a number." That is maybe an answer but it doesn't really help me understand why something as specific as 2.71828... pops up so often and in such important places. – BoddTaxter Aug 09 '16 at 17:36
$D:C^{k} \to C^{k-1}$ given by $D=\frac{d}{dx}$ is just a fancy way of writing $Df=\frac{df}{dx}=f'(x)$. That is hardly jargon, it is treating the derivative as a linear operator, a perspective that is ubiquitous to physics, especially in quantum physics. Again, there is nothing special about $e$ as @Qiaochu Yuan mentioned already. The fact that it is used in finance can be traced to the differential equation $dB(t)=r_{t}B(t)dt$ with initial condition $B(0)=1$ – Wavelet Aug 09 '16 at 17:46
where $B(t)=e^{\int^{t}{0} r{s} ds}$ – Wavelet Aug 09 '16 at 17:47
@BoddTaxter ${d (e^{kx})\over dx} = ke^{kx}$ can you understand this – Aug 09 '16 at 17:48
@ritwiksinha yes. Now I get the the first part of wavelet's answer. To understand the rest I'll have to look into eigenvalues. Currently what I can glean from his answers is that it is special because ${d (e^{kx})\over dx} = ke^{kx}$ which just seems like another example of utility rather than a why. If somebody could start by explaining what, "$e$ is the most natural base." means that would be great. – BoddTaxter Aug 09 '16 at 18:00
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https://www.researchgate.net/post/Why_log_with_base_e_is_frequently_used_as_compared_to_base_10 I'm sorry to say that it is nothing more profound than what I've written above. One could also argue that it makes computations much easier, though again this is a product of what I've written above. – Wavelet Aug 09 '16 at 18:08
Check out Euler's identity too if you haven't already. Ultimately it comes back around to what I've written above but the process of getting there might give you the "math trip" you are looking for. It also makes for a very interesting connection between the Fourier transform and the differentiation operator. – Wavelet Aug 09 '16 at 18:16
@BoddTaxter If you consider this answer a "bunch of jargon", then how did you manage to define $e$ in the first place without using a "bunch of jargon"? Do you have a definition that doesn't involve differentiation, or something equally "jargonish"? If not, what does your question mean? – WillO Aug 09 '16 at 22:15
@WillO I didn't define $e$. I just know it is a number that seems important, more so than similar numbers like 107.532 or whatever. I don't have a definition not involving jargon, in fact that's what I'm looking for. You can take another transcendental like $\pi$ and explain to a farmer that it is the ratio between the diameter and circumference of a circle and he will see why it is so ubiquitous in nature. Wavelet's answer kind of gets at that but Michael Hardy 's answer seems more like it was intended to teach. He also concedes there's a lot more to it rather than "nothing more" – BoddTaxter Aug 10 '16 at 03:37
@BoddTaxter Just out of curiosity (I probably should have asked this first), what is your background? – Wavelet Aug 10 '16 at 21:05
Currently, a wall. – BoddTaxter Aug 11 '16 at 13:12
@BoddTaxter Here's a farmer-friendly explanation of $e$: a farmer is developing a new field. As her tractor passes back and forth over the field a boulder requiring removal is encountered every 22 feet on average. Assuming boulders are placed independently of one another--the presence of a boulder neither increases nor decreases the chance of finding another one nearby--then the probability that a 22 foot stretch is free of boulders is $1/e$. Also the probability of finding exactly one boulder in 22 feet is $1/e$; the probabilities of two, three, or four are $1/(2e)$, $1/(6e)$, $1/(24e)$. – Will Orrick May 28 '23 at 02:40

score 1 · Accepted Answer · answered Aug 09 '16 at 18:21

\begin{align} \frac d {dx} 10^x & = ( 10^x\cdot\text{constant}) \approx 10^x \cdot (2.3) \\[10pt] \frac d {dx} 2^x & = ( 2^x \cdot \text{constant}) \approx 2^x\cdot (0.693) \end{align} etc. It is easy to show that the derivative of an exponential function is a constant multiple of the same exponential function.

Only when the bas is $e$ is the "constant" equal to $1$.

The fact that $x\mapsto e^x$ is its own derivative accounts for its incessant appearance in the study of differential equations. It also accounts for the fact that the "constant" is the base-$e$ logarithm of the base of the epxonential function.

That's the beginning of the story; there's a lot more to it.

The fact that the "constant" is equal to $1$ only when the base is $e$ is analogous to the fact that in the identity $$ \frac d {dx} \sin x = (\text{constant}\cdot \cos x) $$ the "constant" is $1$ only when radians are used rather than some other unit.

Why is $e$ so ubiquitous?

2 Answers2