0
$\begingroup$

I tried hard to not make this duplicate but if I have missed it please flag this question.

So people constantly ask and answer why e is very important, where it came from, and its derivations, etc. "Oh it's so special because it appears here and there, etc."

But what I really want to know is why e has the number 2.718... what is so special about that number? (not the constant!!) Out of all the numbers in the universe why this 2.718... has the property that makes everyone's life easier? (Not asking for the derivation that gives the number)

I don't know if I explained it well or not. Thanks in advance!

Edit: I guess my question is not very well put. The possible duplicate link asks for what is the meaning/origin/significance of e. What I am asking is exactly what the title says. Why did e had to be this number 2.718.. why couldn't any other number say 5.112.. satisfy all the properties and hence be the natural number instead? If you say it's because 5.112.. doesn't satisfy the equations, that is the point. Why 5.112... doesn't but 2.718... does?

$\endgroup$

closed as off-topic by BigbearZzz, Theo Bendit, Eevee Trainer, Moishe Kohan, RRL Apr 4 at 1:27

  • This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ This is absolutely definitely a duplicate. I have seen this question before. I just can't seem to find it. :-) $\endgroup$ – Theo Bendit Apr 4 at 0:26
  • $\begingroup$ @TheoBendit I suspected so! But it's hard to phrase and search out of so many question asking about the importance of the constant (not the number) $\endgroup$ – Seung Apr 4 at 0:27
  • 3
    $\begingroup$ Possible duplicate of Why is Euler's number $2.71828$ and not anything else? $\endgroup$ – Theo Bendit Apr 4 at 0:28
  • $\begingroup$ @TheoBendit I checked that before I posted, but it is quite different from what I am asking.. let me know if I wasn't clear enough $\endgroup$ – Seung Apr 4 at 0:30
  • $\begingroup$ Maybe it's not a perfect match. But, remember that the things that are "special" about a number are the properties it satisfies, and in that way, I think your question matches the summary given at the bottom of the linked question. Maybe you should include some further detail to explain how it's not a duplicate. $\endgroup$ – Theo Bendit Apr 4 at 0:32
2
$\begingroup$

The simple answer is, compute $e$ using one of its many definitions. There are various methods for computing $e$ and finding its decimal expansion up to a given precision. You can do this, and it should agree with digits of $e$ that you know. However, I suspect that this is not the answer you're looking for.

How would you characterise the decimal expansion of $e$? As given in the question, you say $e = 2.1718...$, but I see no pattern here. By what rule can we get the next digit in the expansion? I certainly don't know of one, or at least, I don't know of any rule that doesn't come down to "compute $e$ to a precision of one order of magnitude greater", which essentially takes us back to the simple answer above.

Essentially, the only way you can define this infinite decimal expansion $2.1718...$ properly, without leaning on some unknown pattern that may not be there at all, is by saying "it's the decimal expansion for $e$". Then, the answer to your question is then trivial: the decimal expansion $2.1718...$, which specifically refers to the decimal expansion of $e$, is the decimal expansion of $e$ by definition.

So, unless you can come up with some kind of alternate rule to describe the infinite decimal expansion of $e$, your question is only ever going to have trivial answers.

EDIT: To explicitly address your edit, calculation verifies that $e$ is not approximately any of the other numbers given. For example, using the definition $$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n,$$ we can show that $e \neq 5.112...$, because it can be shown that $$\left(1 + \frac{1}{n}\right)^n \le 3$$ for all $n$, and hence the limit, $e$, must also be less than or equal to $3$. Don't forget, limits are unique!

$\endgroup$
  • $\begingroup$ I am sorry again, but you don't get my point. Could you check my edit? $\endgroup$ – Seung Apr 4 at 1:00
  • $\begingroup$ @Seung I've edited. I'm still not certain what your question is exactly. I still don't see how your question doesn't boil down to "Why this, and not that", which as per my original point, cannot be answered without saying what "this" is. $\endgroup$ – Theo Bendit Apr 4 at 1:05
  • $\begingroup$ Thank you for all the efforts. I ran out of words to explain my thoughts. It could be just a stupid question :( $\endgroup$ – Seung Apr 4 at 1:10
  • $\begingroup$ I may be wrong, but it seems to me that you're aiming for some thunderbolt of clarity to somehow connect the seemingly unremarkable decimal expansion of $e$ with all its various magical properties. If this is indeed what you want, then we simply can't help you. We have no more ability to define the decimal expansion of $e$ than you do, let alone explain some deep significance. $\endgroup$ – Theo Bendit Apr 4 at 1:14
  • $\begingroup$ I think that is what I am asking. Yes, how does the seemingly unremarkable number has all its various magical properties.. I guess it's just what it is. I feel like at this point the concept of my question is too abstract that it could be just a mistuning of my intuition. This will burn my heart for weeks drumming my brain. Finally, to give a context, I began thinking about this after reading about Bernoulli's discovery of e studying compound interest. When n goes infinity for (1+1/n)^n -> 2.718... which is cool I thought, but why 2.718..? $\endgroup$ – Seung Apr 4 at 1:30
1
$\begingroup$

Start with a $\$1$ investment that pays $100\%$ interest per year so that after one year you have $1+1=2$ dollars.

Then suppose instead that you compound twice during the year, once after six months and then again at the end of the year. You would have $(1+\frac{1}{2})^2=\$2.25$.

Notice that, since the interest rate is $100\%$ per year, you only use half of that every six months.

You can do this for an arbitrary number of compounding periods. In general, if we compound $n$ times per year then our interest at each compounding period is $\frac{100\%}{n}=\frac{1}{n}$ and the amount of money you have at the end of the year is $(1+\frac{1}{n})^n$.

A perfectly reasonable question is: how much money will I have if I keep increasing the compounding period. We compute this by taking the limit as $n$ goes to infinity of $(1+\frac{1}{n})^n$. It turns out that this converges to a number and that number is $e\approx 2.718$.

$\endgroup$
  • $\begingroup$ Thanks John, that example is exactly why I started questioning. Your last sentence is the core. "It turns out that this converges to a number and that number is 2.718". It's cool that it converges to a number, but why does it converge to 2.718? $\endgroup$ – Seung Apr 4 at 1:40
  • $\begingroup$ @Seung Once you know it converges then, as others have said, you get an approximation to the actual number by looking at the value for a large $n$. I hope you realize that you will never get an exact number because $e$ is irrational. $\endgroup$ – John Douma Apr 4 at 1:42
  • $\begingroup$ Ah now I get why others got confused. I only write 2.718... because it's irrational and I cannot write all of it. I acknowledge that it cannot be expressed precisely. That is far from my point. Let's call 2.718... "that number". Again, it's cool that it converges to a number, but why does it converge to "that number"? $\endgroup$ – Seung Apr 4 at 1:54
  • $\begingroup$ @Seung Why is $\pi\approx 3.14159$? If you can explain that then maybe I will understand your question better. $\endgroup$ – John Douma Apr 4 at 13:02
  • $\begingroup$ Maybe that is an extension of my quesiton. I cannot answer it. Why is the ratio of circumference to its diameter ≈ 3.1415? $\endgroup$ – Seung Apr 4 at 16:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.