Why do we define $e^x$ as the $\lim_{x \to \infty} (1 + \frac{x}{n})^n$?

Question

I am making my way through an analysis textbook and we have defined $e$ as the $\lim_{n \to \infty} (1 + \frac{1}{n})^n$. We also defined $e^x$ as $\lim_{n \to \infty} (1 + \frac{x}{n})^n$.

I am quite uncertain as to where this exponentiation definition comes from. For example, in the book we have shown how $\sqrt{2}$ can be represented as an infinite decimal. The method we used was to take increasing decimal expansions that were just under $2$, adding one decimal at a time. So for example:

$$1^2 < 2 < 2^2$$ $$1.4^2 < 2 < 1.5^2$$ $$1.41^2 < 2 < 1.42^2$$

and so on. We then define $\sqrt{2}$ to be the supremum of the set of improving approximations of $\sqrt{2}$, sup $\{1, 1.4^2, 1.41^2, 1.414^2,...\}$ and we showed that $2$ is indeed the least upper bound.

So it seems to me that we have already established a way to take irrational numbers to integer powers. So now my question is, why not just do the same for $e^x$? I don't understand why we introduce the limit definition to define exponentiation. Once we are given the definition of $e$, which is just a constant, why can't we use the same procedure as described above to define exponents of $e$? (By same procedure I mean taking increasingly accurate finite decimal approximations of $e$ and multiplying them together $x$ times).

For example, my confusion is along these lines: Say we have defined exponentiation of natural numbers to natural numbers. We define it in the usual intuitive sense, so $3^5 = 3\cdot3\cdot3\cdot3\cdot3$. Then all of a sudden we say introduce a limit to define $7^8$, instead of using the established exponentiation.

I know that for any (integer at least) value of $x$, $\lim_{n \to \infty} (1 + \frac{x}{n})^n$ converges to the same value that we would get if using the other method, so obviously the definition is consistent. However, where exactly does it come from? Do we use it simply because it 'works', and converges to $e^x$ had we used the other method?

Note: I know a caveat to the other method presented here is that in the described form, it can only be used for integer powers. In my book it does say this can be generalised to real powers, but using $e^x$, which doesn't give me much insight.

Answer 1

The definition in your book isn't great. It would make more sense to define the function $e^x = \exp(x)$ as, for example, the unique solution of the differential equation $y' = y$ with $y(0) = 1$. Directly from that definition, it's not hard to prove that $\exp(x + x') = \exp(x) \exp(x')$ (by noting that for fixed $x'$, the function $\exp(x + x')/\exp(x')$ also satisfies the given differential equation), that $e^x$ is defined for the entire real line, that $e^x = \sum x^n/n!$, and so on. With that in mind, the number $e$ is just $\exp(1)$.

More specifically, you asked why we can't define $e^x$ by extending the function $y^x$ for $x, y\in \mathbb{Q}^+$ to real $x, y$ with $y > 0$. You could, but you're not really gaining much from it. You'd have to prove that that extension exists and is well-defined, that it has nice properties like $(yy')^x = y^x (y')^x$ and $y^{x + x'} = y^x y^{x'}$, etc. It's not insurmountable, but it's not as easy as setting $y^x = e^{x \log y}$ and getting most of the properties for free.

As for why $e$ is defined in the text as $\lim_{n\to\infty} (1 + 1/n)^n$, I don't think there's any justification for it aside from being an easy definition to make. It's not motivated, it's not useful, and it's not even interesting.

Answer 2

The difficult is not immediately due to exponentiation, it is due to $e$ being a "special" number which you need to somehow define. In fact, is it transcendental, and there is no way of defining it by simple means, as the root of a polynomial with integer coefficients. (Contrary to $\sqrt2$ which is a root of $x^2-2$.)

The motivation for $e$ is that it is the most "natural" choice for exponentials, as it is the only basis that enjoys the important property

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

More specifically,

$$\lim_{h\to0}\frac{e^{x+h}-e^x}{h}=e^x\lim_{h\to0}\frac{e^{h}-1}{h}$$ and $e$ is the unique number that satisfies the equation

$$\lim_{h\to0}\frac{e^{h}-1}{h}=1.$$

Now if you solve

$$\frac{z^{h}-1}{h}=1,$$ the solution is

$$z=\left(1+h\right)^{1/h},$$ and by setting $h=\dfrac1n$, you get the definition in question.

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Beware that I don't claim that this resolution method is rigorous, you still need to prove that by plugging the last limit in the equation you get $1$.

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Now the definition of the natural exponential easily follows from that of $e$, with

$$e^x=\left(\lim_{x\to 0}\left(1+\frac1n\right)^n\right)^x=\lim_{x\to 0}\left(\left(1+\frac1n\right)^n\right)^x=\lim_{x\to 0}\left(1+\frac1n\right)^{nx},$$ by a continuity argument.

Answer 3

The provided digit-by-digit process you've provided for computing $\sqrt2$ has several drawbacks.

1. It only allows you to define $a^b$ for rational $b$. To get irrational $b$, you would need to go through the whole process again within the exponent.

2. It is hard to work with. It doesn't directly give you what $a^b$ is to work with.

3. It does not obviously have desired exponentiation properties. Does $(a^b)^c=a^{bc}$ and $a^b\cdot a^c=a^{b+c}$? Is it monotone? Continuous? Differentiable?

By playing around with the limit, we can see that this should intuitively give us exponentiation as we know it:

$$\lim_{n\to\infty}\left(1+\frac xn\right)^n\stackrel?=\lim_{n\to\infty}\left[\left(1+\frac xn\right)^{n/x}\right]^x\stackrel?=\lim_{n\to\infty}\left[\left(1+\frac1n\right)^n\right]^x=e^x$$

Of course this is not rigorous seeing as we don't have exponentiation to work with yet, as it is what we are trying to define, but this should help your intuition. It remains to be shown that the listed properties are easily shown, and from them we can prove exponentiation as you know it works.

Answer 4

It's hard to say exactly what you want clarification about. There are at least two things your why may mean (not mutually exclusive, recognise):

(1) What originally motivated this definition of $e^x$? Well, the short answer to this strand is, the problem of continuous compounding of interest. This was originally investigated by one of the Bernoullis -- I've forgotten precisely which at the moment.

(2) Or perhaps you meant to ask why this definition is equivalent to other definitions of $e^x.$ Well, again, a short answer is that all these different definitions satisfy properties that everyone agrees the function $e^x$ should have.

But another possible source of confusion is apparent from your explanation:

So it seems to me that we have already established a way to take irrational numbers to integer powers. So now my question is, why not just do the same for $e^x$? I don't understand why we introduce the limit definition to define exponentiation. Once we are given the definition of $e,$ which is just a constant, why can't we use the same procedure as described above to define exponents of $e$? (By same procedure I mean taking increasingly accurate finite decimal approximations of $e$ and multiplying them together $x$ times).

Well, the short answer to this would be that in the definition you're asking about, $x$ can be any real number whatsoever, not just positive integers as you're suggesting.