Why did the natural constant e had to be 2.718..

Question

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?

Answer 1

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!

Answer 2

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 $e\approx 2.718$.