I'm very eager to know and understand the definition of $e$. Textbooks define $e$ as follows

$$ e = \lim_{p\to\infty} \left[1+\frac1{p}\right]^p \approx 2.71828 $$

Is there an "easy to understand" proof of this? I'm really looking for a derivation of this which is very intuitive and easy to comprehend.

By the way I'm watching this video lecture.

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    $\begingroup$ You can't prove a definition. That doesn't really make sense. $\endgroup$ Sep 3, 2011 at 18:06
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    $\begingroup$ Then where did this come from?There has to be a possible derivation of this. $\endgroup$
    – alok
    Sep 3, 2011 at 18:10
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    $\begingroup$ Related... $\endgroup$ Sep 3, 2011 at 18:11
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    $\begingroup$ ... and this ... and this $\endgroup$
    – Henry
    Sep 3, 2011 at 18:21
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    $\begingroup$ Raskolnikov is right: you can't prove a definition. You can prove, however, that this is a good definition, i.e. that the limit exists, and is finite. $\endgroup$
    – zar
    Sep 3, 2011 at 18:58

3 Answers 3


What is your definition of $e$? That's the main problem. One cannot prove a definition, though one can prove that several different ways of defining something yield the same thing.

There are several "roads to $e$". One road is via derivatives:

What continuous, differentiable function $f(x)$ satisfies $f'(x) = f(x)$ for all $x$, and $f(0)=1$?

This is the problem that Euler considered. If you think functions are given by Taylor series, then starting from $$f(x) = 1 + a_1x + a_2x^2 + a_3x^3 + \cdots + a_nx^n + \cdots$$ and taking derivatives formally, $$f'(x) = a_1 + 2a_2x + 3a_3x^2 + \cdots + na_nx^{n-1}+\cdots$$ then identifying coefficients you get $$a_1 = 1,\quad a_2 = \frac{a_1}{2},\quad a_3 = \frac{a_2}{3}=\frac{a_1}{3!},\ldots$$ so that the function in question is $$f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots$$ and $e$ happens to be the value of this function at $1$. In fact, since the function has $f'(x)/f(x)$ constant, it is an exponential function, so $f(x) = e^x$.

Is there something intuitive about this? Perhaps; if you compare the growth of, say, $2^x$ and $3^x$, you will see that $2^x$ grows a little slower than its value ($f'(x)\lt f(x)$), whereas $3^x$ grows a little faster $(f'(x)\gt f(x))$. Somwhere in between, you expect to find a function which grows exactly the same as its value, and that "somewhere" is precisely at $e$.

This is a somewhat "continuous" definition of $e$: you arrive at $e$ by thinking about a continuous process (functions and differentiation).

There is a different approach, one which is a bit more "discrete."

Imagine putting money in a bank at interest. How much money you have at the end of the year depends on how often you get paid your interest, because interest paid will itself generate interest. (Aside: that's why in your Certificates of Deposit and Credit Card statements there is a difference between the "Interest Rate" and the "APR" or annual percentage rate; one is the nominal interest you get paid, the other one is how much you actually get due to compounding).

Let's imagine that you put \$100 at 100% annual interest. If you get paid interest at the end of the year, then you simply get \$100 interest at the end of the year, so your total is $$ 100+ 100(1)= 100(1+1) = 200\text{ dollars.}$$

Now say you get paid interest every six months instead. After six months, you've earned half the interest you were entitled to, \$50. So now you have \$150 in the bank for six months, which earns you half of the 100% you would get for a whole year, or \$75. So at the end of the year you have: $$100 + 100\left(\frac{1}{2}\right) + \left(100 + 100\left(\frac{1}{2}\right)\right)\frac{1}{2} = 100\left(1 + \frac{1}{2}\right)^2 = 225\text{ dollars.}$$ If you are paid every four months, then you get $\frac{1}{3}$ of the interest, three times, so at the end of the year you have (if you do the calculations and simplify using the formula for a geometric sum): $$100\left( 1+ \frac{1}{3}\right)^3 \approx 237.04\text{ dollars.}$$ If you are paid every month, at the end of the year you will have $$100\left(1+\frac{1}{12}\right)^{12} \approx 261.30\text{ dollars.}$$ If you are paid every day, at the end of the year you will have $$100\left(1 + \frac{1}{365}\right)^{365} \approx 271.46\text{ dollars.}$$ If you get paid every hour, at the end the year you will have $$100\left(1 + \frac{1}{8760}\right)^{8760} \approx 271.81\text{ dollars.}$$

And so on. If you get paid every minute you'll get more; every second, even more. Every nanosecond, even more. And so on. The question is:

Suppose you keep making the intervals at which you are paid smaller and smaller and smaller. At "the limit", when you are being paid interest every single instant, will you become infinitely rich?

No. It turns out that the sequence $$100\left(1 + \frac{1}{n}\right)^n$$ is strictly increasing, but bounded above (by, for example, \$300). That means that the sequence converges: it approaches a definite, specific number. What is that number?

It's $100e$. One can prove that the limit you get by doing $$\lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n$$ is exactly the same as the base of the function we defined above by asking for a differentiable function that equals its derivative and has value $1$ at $0$. So this is another way of defining $e$.

You can take either one as the basis for defining $e$, and then prove that the other construction gives you the same thing. You pick one, you prove the other. Which one is more "intuitive"? Well, in a way neither one is intuitive: it just happens to be the answer to a specific question; perhaps a better way to phrase your question is: "What is a reasonable question we might ask where the answer is $e$?" and now you have two such questions.

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    $\begingroup$ This is a very nicely written answer. $\endgroup$
    – davidlowryduda
    Sep 3, 2011 at 19:20
  • $\begingroup$ @Arturo:Sexy answer! I'am gonna show this to my mom and whomever i know.I don't think i need to know much more about the origin of e atleast for the time being.This rational explanation will suffice. $\endgroup$
    – alok
    Sep 3, 2011 at 20:48
  • $\begingroup$ I wanted to revisit the definition of $e$ and couldn't resist leaving a comment here because this is such a wonderfull explanation.It's beautifull as $e$ itself! $\endgroup$
    – alok
    Dec 12, 2011 at 18:01

$\pi$ is the name of the constant relating the diameter and circumferance of a circle. It's a definition, a particular constant that we thought deserved a name. $e$ happens to be the name of a constant from a particular limit. Like $\pi$, we named it because we thought it would be useful.

In Physics, many, many constants have names. Gravitational constants, expansional constants, electrical constants, etc. Each is useful, but somewhat arbitrarily chosen.

But I suspect you won't like the apparently nature of this. So I come up with something related: We call 2 the number s.t. 1 + 1 = 2. Why? What is its derivation? There is no derivation - we defined it, gave it a symbol, and a name.

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    $\begingroup$ Your explanation of constants in general and $e$ in particular is beautifull too. $\endgroup$
    – alok
    Dec 12, 2011 at 18:04

It's really easy to show that $\dfrac{d}{dx} 2^x = (2^x\cdot\text{some constant})$, and that $\dfrac{d}{dx} 4^x = (4^x\cdot\text{some other constant})$.

At http://wnk.hamline.edu/~mjhardy/1170/homework/6th.pdf [later edit: Here's another (actually older) version, the link to which is still working: http://www.math.umn.edu/~hardy/1271/handouts/September.28.pdf] you will see in problem #2 that it's really easy to see that with 2, the "constant" is less than 1, and with 4, the constant is more than 1.

By this point, you'll be suspecting that for some number somewhere between 2 and 4, the "constant" will be 1.

That number is $e$.

That's one way to begin understanding what $e$ is about.

Now recall that the derivative of $f$ is approximated by $\dfrac{f(x+h)-f(x)}{h}$ when $h$ is small. Think about what happens when $h = 1/n$, and that may shed some light on $\left(1 + \dfrac{1}{n}\right)^n$.


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