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.

You have a typo: $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ — PinkyWay, Feb 20 '20 at 13:23
"Once we are given the definition of $e$,..." And how do you have $e$ defined? Most commonly one would define $e$ as $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ or as $\text{exp}(1)=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots$. It is hard to define $e$ without also touching upon $e^x$ at the same time. — JMoravitz, Feb 20 '20 at 13:26
@JMoravitz yes I realise that it's a bit arbitrary to complain about $e^x$ when $e$ seems to appear magically as well. I think with $e$ I am less concerned because in some way we are defining a constant (albeit with particularly important properties). Whereas for exponentiation, it seems to deviate from definitions/approaches I have already seen, and I am not sure why. — masiewpao, Feb 20 '20 at 13:28
@masiewpao It doesn't deviate at all. It's just that you haven't worked out the equivalence between your intuition and the definition (hint: there a million questions showing that equivalence on this site, and its not hard to do yourself either) — Rushabh Mehta, Feb 20 '20 at 13:32
$$e^x = \lim_{n \to \infty} (1 + \frac{1}{n})^{nx}$$ take $m = nx$ $$e^x = \lim_{m \to \infty} (1 + \frac{x}{m})^{m}$$ equivalently $$e^x = \lim_{n \to \infty} (1 + \frac{x}{n})^{n}$$ — Aven Desta, Feb 20 '20 at 13:45
@DonThousand Do you mean they are equivalent definitions? I read through a number of questions but for some reason I couldn't 'get' it. But the answers below made me realise my approach was wrong, in that I was getting stuck on why the definition seemed to come from nowhere, rather than think about fitting the definition to the desired properties. — masiewpao, Feb 20 '20 at 13:58
How do you define "multiplying them together $x$ times" when $x$ is not a positive integer? Defining the symbol $a^b$ when $b$ is irrational is a difficult problem and can't be handled by algebraic methods. Most of the intuitive approaches are quite complicated. It is much easier to use the definition $e^x=\lim_{n\to\infty} (1+(x/n))^n$. — Paramanand Singh, Feb 20 '20 at 17:23
Also there are many routes to the development of theory of exponential and logarithmic functions and I have discussed them at length in my blog (search "exponential" in archives) and also on MSE. — Paramanand Singh, Feb 20 '20 at 17:39

score 2 · Answer 1 · answered Feb 20 '20 at 13:43

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

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

score 1 · Answer 2 · 2020-02-20T13:42:19.103

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.

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 $1$.

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.

score 0 · Accepted Answer · answered Feb 20 '20 at 13:50

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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.

answered Feb 20 '20 at 13:50

anomaly

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Not motivated or interesting? Historically, it's actually one of the first appearances of the number $e$ (coming from the study of compound interest); see for example http://mathshistory.st-andrews.ac.uk/HistTopics/e.html. – Hans Lundmark Feb 20 '20 at 14:23
@HansLundmark: Sure, that's the explanation that usually gets trotted out to justify the definition, but I wouldn't consider it a particularly compelling motivation. As a practical matter, if your bank offers anything close to continually compounded interest or even an interest rate where that sort of approximation is useful (and you have a calculator that can compute exponentials but not $(1 + 1/n)^n$), it's better than mine. As a less practical matter, there are more interesting early applications of the exponential function: normal distributions, various discrete probability problems, etc. – anomaly Feb 20 '20 at 14:30
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I'm not advocating that this definition should be used nowadays. The exponential function is more fundamental than the number $e$, and I agree with you that the easiest definition is via the ODE. I just thought it was a bit categorical to totally dismiss it, since it is at least of historical interest. – Hans Lundmark Feb 20 '20 at 14:55
Almost great answer. Albeit spoilt by your opinionated last sentence, driven by ignorance. FYI the form $1+\epsilon,$ where $\epsilon$ is a small positive number, is a very important one in analysis. Also, this definition arose from investigations into continuous compounding of interest in a given time interval. – Allawonder Feb 20 '20 at 17:05
The definition of $e^x$ as limit of $(1+(x/n))^n$ is much simpler than other approaches as the proofs are simpler. The approach you mention requires some effort to show the existence and uniqueness of a solution of differential equation and it ultimately brings the logarithm as an integral in picture. – Paramanand Singh Feb 20 '20 at 17:27
Having read through the answers and comments I feel like this answer made it click for me, but maybe I should clarify. What you are saying is essentially that we define the exponentiation in a way (any way), such that we have some desired properties correct? For some stupid reason I thought the definition came from nowhere, and I wondered why these wonderful properties popped out, rather than realising that in reality it was the reverse; such definitions are crafted to satisfy these properties on purpose. – masiewpao Feb 20 '20 at 17:34
@Allawonder: Don't be rude. – anomaly Feb 20 '20 at 17:36
@masiewpao: Think of it like $\pi$. Historically, it has the clear motivation of being the ratio of the circumference of a circle to its diameter. That's a perfectly reasonable definition, but $\pi$ has since become ubiquitous throughout mathematics, even in areas that are (ostensibly, at least) unrelated to plane geometry. You can define $\pi$ via a variety of infinite sums, or probabilistic processes, etc.; and even all those definitions are equivalent, they make some things easier to prove and others harder, and some of the definitions are more compelling than the others. – anomaly Feb 20 '20 at 17:42
@masiewpao: those definitions are not crafted but arise naturally during analytical investigation. For example if you start with functional equation $f(x+y) =f(x) f(y) $ you immediately get the equation $f'(x) =f'(0)f(x)$ which brings us to the approach mentioned in this answer. For a slightly crafted approach see this thread. – Paramanand Singh Feb 20 '20 at 17:45
@anomaly thank you, that's a very good analogy. After thinking about your answer I think that is exactly how I thought of the definitions for $e$, especially since I had read a question on here about the various ways it can be defined. I am really not quite sure why I was confused initially, it doesn't really make sense. – masiewpao Feb 20 '20 at 17:48
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@masiewpao: Happy to help! – anomaly Feb 20 '20 at 17:51
@ParamanandSingh I'm sorry, I'm not sure I understand what you mean. I think my use of the word crafted was possibly wrong. Wouldn't whatever definition/characterisation we use need to be defined such that certain properties are satisfied? And that is why we can end up with so many different but valid definitions? – masiewpao Feb 20 '20 at 18:00
@anomaly Thanks for the anomalous reminder, but I promise I don't intend to be rude. – Allawonder Feb 20 '20 at 18:27
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@masiewpao: Perhaps I was not clear enough. If we assume properties of exponential functions then these assumptions lead us to the definitions. It was not like someone was trying to find definitions which guarantee the properties. – Paramanand Singh Feb 21 '20 at 02:21
@ParamanandSingh Ah I get you, that makes sense to me now thank you! – masiewpao Feb 21 '20 at 15:19

score -1 · Answer 4 · answered Feb 20 '20 at 16:57

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.

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

4 Answers4