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Why is Euler's number $\mathtt 2.71828$ and not for example $\mathtt 3.7589$?

I know that $e$ is the base of natural logarithms. I know about areas on hyperbola xy=1 and I know its formula: $$e =\sum_{n=0}^\infty \frac{1}{n!} \approx 2.71828$$ And I also know it has many other characterizations. But, why is $e$ equal to that formula (which sum is approximately $\mathtt 2.71828$)?

I googled that many times and every time it ends in having "$e$ is the base of natural logarithms". I don't want to work out any equations using $e$ without understanding it perfectly.

Summary: I'm looking for the origin of $e$, if $\pi$ came from the radius of a circle with a unit diameter, then what is $e$ ???

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    $\begingroup$ Because that's that what the summation sums up to?... $\endgroup$ – user635162 Jan 29 at 16:05
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    $\begingroup$ It depends how you define $e$. If you define $e$ as the sum, then it's $2.718\dots$ because that's what the sum equals. If you define it as the limit of $(1+1/n)^n$, then it's $2.718\dots$ because that's what the limit equals. $\endgroup$ – Wojowu Jan 29 at 16:06
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    $\begingroup$ "I don't want to work out any equations using e without understanding it perfectly." John von Neumann, possibly one of the greatest mathematical and scientific minds of the modern era, said "In mathematics you don't understand things. You just get used to them." $\endgroup$ – icurays1 Jan 29 at 16:07
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    $\begingroup$ Do you also wonder why $\pi$ is $3.1415....$? $\endgroup$ – Randall Jan 29 at 16:07
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    $\begingroup$ Isn't Euler's constant $0.5772156649$? $\endgroup$ – YuiTo Cheng Jan 29 at 16:08
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$\sum\frac1{n!}$ is not that special.

$\lim_{n\to\infty}\left(1+\frac1n\right)^n$ is not really special.

$f'(x)=f(x)$ is a very simple differential equation, but unremarkable, really.

$\ln (x)$ is only marginally nicer than other logarithms, in that its derivative is $\frac1x$.

The fact that a single number connects all of these (and many, many others) as intimately as $e$ does is nothing short of a miracle. Oh, and also $e$ happens to have the decimal expansion $2.718\ldots$

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    $\begingroup$ Well summarized! $\endgroup$ – Thomas Lesgourgues Jan 29 at 16:10
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    $\begingroup$ @YvesDaoust Yes, you're right. $2$ is a very important mathematical constant too. I'm not saying it's a competition for which constant is best. But the number $e$ appears in so many contexts that not having a name for it is counterproductive. So therefore it has been given the name $e$. $\endgroup$ – Arthur Jan 29 at 16:36
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    $\begingroup$ @YvesDaoust What do you mean we didn't? $2$ is a pretty important constant too! $\endgroup$ – Wojowu Jan 29 at 16:36
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    $\begingroup$ So maybe the real question is why sequences of reciprocals of factorials pop up all over the place. $\endgroup$ – timtfj Jan 29 at 16:38
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    $\begingroup$ Of course if you dig deeper, you can come up with connections between these things. For example, if you apply Euler's method with $n$ steps to find a numerical approximation to a solution to $f'(x)=f(x), f(0)=1$ then that approximate solution has $f(1)=(1+\frac{1}{n})^n$. $\endgroup$ – Daniel Schepler Jan 30 at 0:27
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We use $e$ because it a natural choice, as it yields a simple derivative:

$$(e^x)'=e^x.$$

For other bases, we have

$$(a^x)'=\ln a\,a^x$$ and the factor $\ln a$ is annoying.

For a very similar reason we use radians in the trigonometric functions:

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

With degrees, we would have

$$(\sin_d x)'=\frac\pi{180}\cos_d x,$$ once more an embarrassing factor.

As shown by Hyperion, the condition $(e^x)'=e^x$ induces the value

$$1+1+\frac12+\frac1{3!}+\frac1{4!}+\cdots$$


Assume you wanted to find a number $b$ such that $(b^x)'=b^x$. Using the definition of the derivative, you could try to solve

$$\frac{b^{x+h}-b^x}h\approx b^x$$ where $h$ is a small increment.

Then $$\frac{b^{x+h}-b^x}h=b^x\frac{b^h-1}h\approx b^x$$ leads to

$$b^h\approx 1+h$$ or $$b\approx(1+h)^{1/h}.$$

It turns out that this expression has a limit for $h\to0$, which you can obtain using the generalized binomial theorem.

E.g.,

$$1.000001^{1000000}=2.718280469\cdots$$

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Clearly, one answer is "because that's the value that the various definitions produce, and when we follow them $\sum_{n=0}^{\infty}\frac{1}{n!}$ pops out". But it's not a very satisfying answer (in fact I think you're asking for an underlying reason why that happens).

I can't give a definitive why, but my suggestion is that it's something to do with iterated processes like

  • taking the next derivative
  • dividing by the next integer
  • choosing the next item in a permutation
  • multiplying by the next bracketed expression

all of which are quite good at producing sequences of factorials.

But of course I've now got $e^{iπ}=-1$ nagging at me, and even though that can be explained in terms of "exponential growth sideways" and proved to be true, it doesn't in itself seem that related to any iterated process, and @Arthur's comment that it's "nothing short of miraculous" seems more accurate than any proof of the connection would be.

My suggested explanation, if true, just pushes the question back a level: "Why do iterated processes that produce the series for $e$ pop up all over the place?"

Typically if you ask Why? more than about four or maybe five times (following underlying reasons rather than a chain of trivial causal events or a string of theorems), you'll get to unanswerable philosophical questions—for instance "Why is it raining?" leads me after a few steps to "why is there such a thing as the laws of physics?" I suspect that pursuing the reasons why a particular number is as it is will have the same result.

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    $\begingroup$ How do you not see exponential growth as an iterative process, given the standard definition of exponentiation as repeated multiplication of the same number? $\endgroup$ – Nij Jan 29 at 23:41
  • $\begingroup$ @Nij I do, but I mean that $e^{iπ}=-1$ doesn't intuitively seem like an example of it—it looks more like an amazing static relationship between $π$, $i$ and $e$ which can be proved by talking about exponential growth but is really a free-standing thing of its own. $\endgroup$ – timtfj Jan 30 at 0:17
  • $\begingroup$ @timtfj But Euler's identity (of which the equation above is only a product) is related to an iterative process, namely that of repeatedly multiplying an arbitrary complex number by some complex $z.$ $\endgroup$ – Allawonder Jan 31 at 16:55
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    $\begingroup$ @Allawonder of course, but the point is that it's not necessarily intuitive to see it that way—in fact some people struggle hard to. The problem is that logic of the situation seems to be at a surface level, distinct from the level of the WHY??? that people are asking themselves. As though the logical explanation amounts to the mere mechanics of something, or to describing the features of something that already exists rather than explaining why it exists. I think it's really a subjective or philosophical issue, not a mathematical one. $\endgroup$ – timtfj Jan 31 at 17:11
  • $\begingroup$ @timtfj I wasn't trying to explain why $e$ appears to be connected with iterated operations; I was only pointing out that contrary to what you thought, the complex exponential defines one of the most intuitive iterative operations. $\endgroup$ – Allawonder Jan 31 at 17:22
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Why is Euler's number 2.718 and not anything else?

Short answer: by definition so.

First paragraph of the Wikipedia article $e$ (mathematical constant):

The number $e$ is a mathematical constant that is the base of the natural logarithm: the unique number whose natural logarithm is equal to one. It is approximately equal to $2.71828$, and is the limit of $(1 + 1/n)^n$ as $n$ approaches infinity, an expression that arises in the study of compound interest.

... why is $e$ equal to that formula (which sum is approximately $𝟸.71828$)?

"That formula" is one of the equivalent definitions of the constant $e$. All the equivalent definitions has the same approximate value $𝟸.71828$.

I googled that many times and every time it ends in having "e is the base of natural logarithms". I don't want to work out any equations using e without understanding it perfectly.

Should you have any similar question in the future, the first thing you should ask is what is the definition of the mathematical object that you are confused about.


For history of the constant $e$:

https://en.wikipedia.org/wiki/E_(mathematical_constant)#History


[Added to respond to a comment below.]

The way you phrase your question is problematic. The constant $e$ is not discovered by mathematicians. It is defined to be the constant $\lim_{n\to\infty}(1+\frac{1}{n})^n$, which has the approximate value $2.71828$. What mathematician do is nothing but give an interesting constant a name. If Bob calls his dog "Alpha", it does not make much sense to ask "Why is Alpha a dog, not a cat?" --- because Bobs calls his dog "Alpha"!

On the other hand, it is reasonable to ask what the "story" about $e$ is, where it appears and why it is interesting. I believe this is what you really wanted to ask.

You may want to take a look at this article:

An Intuitive Guide To Exponential Functions $\&$ $e$

Here is an excerpt:

Describing e as “a constant approximately 2.71828…” is like calling pi “an irrational number, approximately equal to 3.1415…”. Sure, it’s true, but you completely missed the point.

Pi is the ratio between circumference and diameter shared by all circles. It is a fundamental ratio inherent in all circles and therefore impacts any calculation of circumference, area, volume, and surface area for circles, spheres, cylinders, and so on. Pi is important and shows all circles are related, not to mention the trigonometric functions derived from circles (sin, cos, tan).

e is the base rate of growth shared by all continually growing processes. e lets you take a simple growth rate (where all change happens at the end of the year) and find the impact of compound, continuous growth, where every nanosecond (or faster) you are growing just a little bit.

e shows up whenever systems grow exponentially and continuously: population, radioactive decay, interest calculations, and more. Even jagged systems that don’t grow smoothly can be approximated by e.

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    $\begingroup$ This answer hits the nail on the head. Such a question would rather fit on a philosophy-site. $\endgroup$ – Peter Feb 1 at 10:27
  • $\begingroup$ Thank you very much, but as mentioned above (in the question) : "I googled that many times", I (of course) have seen Wikipedia's article about $e$ (which you shared above), and it doesn't help me to recognize the origin of $e$. I think the mathematicians found out the metioned things on Wikipedia AFTER discovering $e$. so, I'm looking for what they were searching for before finding $e$. $\endgroup$ – anas pcpro Feb 2 at 14:36
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    $\begingroup$ @anaspcpro: You asked an interesting topic in a confusing way. NOT AT ALL "after", please read the History part of the Wikipedia article: en.wikipedia.org/wiki/E_(mathematical_constant)#History $\endgroup$ – user587192 Feb 2 at 15:43
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We can derive that formula through the use of Maclaurin series. If you are unsure of what a Maclaurin series is at this moment of time, it a a method of representing any function in a certain interval as an 'infinite polynomial'. The general formula for the Maclaurin series for $f(x) = e^x$ is $$f(x) = e^x = f(0) + f'(0)x + f''(0)\frac{x^2}{2!} + f''(0)\frac{x^3}{3!} + ...$$ Because the derivative of $e^x$ is equal to itself, plugging in $1$ to the infinite series, we find that $$e^1 = e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ...$$

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    $\begingroup$ This is a proof for the formula which doesn't answer my question $\endgroup$ – anas pcpro Jan 29 at 17:00
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    $\begingroup$ @anaspcpro you asked why $e = \sum_{n=0}^{\infty} \frac{1}{n!}$, and I showed you where it is derived from. $\endgroup$ – Hyperion Jan 29 at 17:03
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    $\begingroup$ yes, but, what I was searching for is "what was mathematicians searching for and brought up e ????" $\endgroup$ – anas pcpro Jan 29 at 17:16
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    $\begingroup$ @Arthur showed what I was looking for... and your answer helped me understand what he said :). Thanks a lot. $\endgroup$ – anas pcpro Jan 29 at 17:21
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    $\begingroup$ @anaspcpro That wasn't at all obvious from your question. Could you edit your question to make it clearer? "What were mathematicians searching for when they 'invented' e?" is a completely different question to "Why is e=2.718...?" $\endgroup$ – David Richerby Jan 29 at 17:31
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No, the reason is that putting 1 dollar to bank with 100% interest rate you will get 2 dollars after 1 year. That is very simple and it is about constant 2.

But, if you will want more, you will put there your dollar for half a year, you will get 1.5 dollar, then you will put this amount to bank again and after another half a year you will have 2.25 dollars.

Increasing the frequency of put - get you will end up with 2.718...$ which is the constant e.

Also notice that exp'(0) == 1 x'(0) == 1 i.e. at 0 they have the same growth.

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    $\begingroup$ Good example. Though this is not how real banks calculate real interest (they know what will happen if they calculated fractional year interest like you suggested) $\endgroup$ – lalala Jan 29 at 19:05
  • $\begingroup$ @lalala most banks I know of calculate fractional year interest, typically 1/12, 1/4, or 1/2 of a year depending on the account type. Some accounts only pay interest annually, but in my experience, monthly or quarterly are the most common. $\endgroup$ – Stobor Jan 30 at 4:28
  • $\begingroup$ Please understand that e is about infinitesimal linear growth with slope 1. $\endgroup$ – Přemysl Nedvěd Jan 31 at 15:45
  • $\begingroup$ Please understand that e is result of infinitesimal linear growth with slope 1 on interval [0;1]. An example of this is the bank account mentioned above. All definitions using series or logarithms are mathematically correct but not direct. $\endgroup$ – Přemysl Nedvěd Jan 31 at 15:53
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    $\begingroup$ This is another characterization for $e$ which I've found on Wikipedia, but didn't help so much.... $\endgroup$ – anas pcpro Feb 2 at 14:41
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One introduction of $e$ can be seen as a solution to $f'(x)=f(x)$. Indeed with $\lambda$ any constant, the function $$ f \ : \ x \mapsto \lambda e^x$$ is so that at each point, its "increase" equals its value, hence $f'(x)=f(x)$

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For example, because the equally perplexing expansions

$$\cos x = \sum^{\infty}_{n=0} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots $$ $$\sin x = \sum^{\infty}_{n=0} (-1)^n\frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots$$

are linked by

$$e^{ix} = \cos x + i\sin x$$

and they all hold too when $x=-i$, leading to your formula when you insert the first two in the third one. When $x=\pi$ you have the beautiful Euler's identity: $e^{i\pi}=-1$.

So, in this sense, your question is an understated invitation to contemplate an apparition of mathematical beauty and the uses of it.

Attaching numerical estimates to these constructs is a possibility that would be unreasonable to dismiss a priori, for quantification is a fundamental method of enquiry and, after all, of finding our our own way in this world. The answer to the question 'what is this?' is completed by the answer to the question 'how much is this?'.

It may take a long time before the penny drops, but the simplicity of some formulas may be flabbergasting on second thoughts. By no chance, by the way, transcendental numbers such $e$ have attracted the attribution trascendental.

Proceed safely and keep on wondering.


Approaching your question

Why is Euler's number 2.71828 and not for example 3.7589?

more literally, the definition of $e$ you quote also reveals the structure of this number (something some numbers do not seem to have, so plain are they). Since a number is made of digits, I have compiled a table where you can see which terms in the expansion contribute to each digit of $e$, restricting myself to the first 8 decimal digits. So you can see the process by which summing successive terms adds information the value of $e$. The tabulation is not entirely foolproof, subject to false negatives occurring, but gives an idea.

enter image description here

Reconnecting to the first part of this answer, I like to picture to myself that the 2 in $e$ is bred by $\cos 0 = 1$ and $-i^2=1$ (a trigonometry because and a complex-number because) and this cannot sum up to 3 (the arithmetics because). One can $\infty$-ly keep on elaborating on these lines and add more and more layers of considerations as you go.

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